Questions on Algebra: Radicals -- complicated equations involving roots answered by real tutors!

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Question 972558: how do you simplify radical 8x squared
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
how do you simplify radical 8x squared
~~~~~~~~~~~~~~~~~~~~~~~~


        @lwsshak3 gives the answer in his post 2x%2Asqrt%282%29.

        This answer is incorrect, and I will explain it below.


In expression  sqrt%288x%5E2%29,  value of x can be positive or negative (or zero).

Expression  sqrt%288x%5E2%29  makes sense and is defined for both positive and/or negative values of x.

Also, according to the common agreement, square root of a number is understood in a middle school as a positive value.

THEREFORE,  the correct answer to the problem's question is   2%2Aabs%28x%29%2Asqrt%282%29.

It works universally for positive and negative values of  'x',
while expression  2x%2Asqrt%282%29  gives a negative value for negative  'x',  which contradicts to the common agreement.

This question/problem is a standard  highlight%28highlight%28TRAP%29%29,  and almost all
unfortunate newcomers, who are not familiar with this explanation,  fall into this trap.


The percentage of these unfortunate newcomers is about  99.999%,  I think,  or even more than that.




Question 163137: Hi, this is the problem I was told to do:
[SQRT(2x + 2)] - [SQRT(x - 3)] = 2
or
√(2x+2) - √(x-3) = 2
I've been told to FOIL this problem, but I don't understand how to FOIL it if its subtracting and not multiplying. What I do know is that the answer is equal to 7.
Thanks in advance.

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, this is the problem I was told to do:

[SQRT(2x + 2)] - [SQRT(x - 3)] = 2
or
√(2x+2) - √(x-3) = 2

I've been told to FOIL this problem, but I don't understand how to FOIL it if its subtracting and not
multiplying. What I do know is that the answer is equal to 7.
Thanks in advance.
*****************
      sqrt%282x+%2B+2%29+-+sqrt%28x+-+3%29+=+2
This might be easier and less confusing if one of the radicals is moved to the right side. So, let's choose -+sqrt%28x+-+3%29 to do that with. 
                    sqrt%282x+%2B+2%29+=+2+%2B+sqrt%28x+-+3%29 ----- Adding  sqrt%28x+-+3%29 to both sides
               %28sqrt%282x+%2B+2%29%29%5E2+=+%282+%2B+sqrt%28x+-+3%29%29%5E2 -- Squaring both sides
               %28sqrt%282x+%2B+2%29%29%5E2+=+%282+%2B+sqrt%28x+-+3%29%29%282+%2B+sqrt%28x+-+3%29%29 
                        2x+%2B+2+=+4+%2B+2sqrt%28x+-+3%29+%2B+2sqrt%28x+-+3%29+%2B+%28sqrt%28x+-+3%29%29%5E2 ---- FOILing right-side
                        2x+%2B+2+=+4+%2B+4sqrt%28x+-+3%29+%2B+x+-+3
                        2x+%2B+2+=+1+%2B+4sqrt%28x+-+3%29+%2B+x
             2x+%2B+2+-+1+-+x+=+4sqrt%28x+-+3%29
                          x+%2B+1+=+4sqrt%28x+-+3%29
                      %28x+%2B+1%29%5E2+=+%284sqrt%28x+-+3%29%29%5E2 ---- Squaring both sides
                 x%5E2+%2B+2x+%2B+1+=+16%28x+-+3%29
                 x%5E2+%2B+2x+%2B+1+=+16x+-+48
x%5E2+%2B+2x+%2B+1+-+16x+%2B+48+=+0
            x%5E2+-+14x+%2B+49+=+0
                     %28x+-+7%29%5E2+=+0
                          x - 7 = 0___x = 7

             You can do the CHECK!!


Question 720317: Here is my problem.
[SQRT(x + 7)] - 2[SQRT(x)] =-2
OR
√(x + 7) - 2√(x) = -2
-------
Now, the steps I took to solve this problem are to first square both sides:
[√(x + 7) - 2√(x)] * [√(x + 7) - 2√(x)] = 4
so then I FOIL the left side, resulting in:
(x+7) - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] + 4x
So then I thought to subtract (x+7) and (4x) to both sides
- [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] = 4 - 4x - x - 7 (I think I'm supposed to switch the sign, because I've subtracted it and moved it to the opposite side, right?)
I think I'm correct up to this point, but now I have to square both sides again.
I think this left hand side could be re-written as:
-2[√(x + 7) * - 2√(x)]
Is this right? I'm subtracting it from itself, a negative, which could simply multiplied by -2. Anyway, now I need to square this again, so I assume the -2 becomes a 4 and I FOIL them separately?
FOILING the left side will get:
[√(x + 7) * - 2√(x)] * [√(x + 7) * - 2√(x)]
Which, when FOILed, looks like
(x+7) - [√(x+7) * -2√(x)] - [ - 2√(x) * √(x + 7)] + 4x
It looks exactly the same as before!! I'm just really confused by this problem, and I have a couple more like it, so I want to know if figuring this one out could help me solve the other ones.
I'm confused about FOILing the different sides, whether or not I can combine two square roots, and quite frankly, a lot of other things.
One of the options on the test is 9, and I think this is the answer, because I've inserted it into the original equation and it works, but I'm just confused about how to actually get 9 out of this..
Sorry for the long question. I hope it's not hard to understand. I'm just hoping someone can walk me through all the steps of solving a problem like this so I can do it easily in the future.

Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Your original equation is sqrt%28x%2B7%29-2sqrt%28x%29=-2

You CAN start the problem by squaring both sides of the equation; but when you do that you end up with an equation involving an "uglier" square root. Reaching the final answer is almost always easier if you start by changing the equation so that there is only one radical on each side of the equation.

And in this particular example, with the "-2" on the right side, I would also, to avoid possible future confusion, move that term to the left hand side, making the equation

sqrt%28x%2B7%29%2B2=2sqrt%28x%29

Now square both sides.

%28x%2B7%29%2B4sqrt%28x%2B7%29%2B4=4x

Now isolate the remaining radical and square both sides again.

4sqrt%28x%2B7%29=3x-11
16%28x%2B7%29=9x%5E2-66x%2B121
16x%2B112=9x%5E2-66x%2B121
9x%5E2-82x%2B9=0
%289x-1%29%28x-9%29=0

x=1%2F9 or x=9

We squared both sides of the equation in solving the problem, so some of the roots we ended up with might not satisfy the original equation, so we need t check. x = 1/9 does NOT satisfy the original equation; x = 9 does. So the unique solution to the given equation is x = 9.

ANSWER: x = 9



A final note....

I work with high school math students who often take timed competitive tests where the speed of getting the answer is important. While the intent here is almost certainly to solve the problem by formal mathematical methods, the problem can be solved informally VERY quickly using logical reasoning.

The original equation shows that a combination via addition or subtraction of the radicals sqrt(x+7) and sqrt(x) yields in an integer result. That means (x+7) and (x) must both be integer perfect squares. A little knowledge of perfect square integers immediately tells us that the only two perfect square integers that differ by 7 are 16 and 9.

So we quickly have our solution:
ANSWER: x = 9


Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Here is my problem.

[SQRT(x + 7)] - 2[SQRT(x)] =-2   OR   √(x + 7) - 2√(x) = -2 
-------
Now, the steps I took to solve this problem are to first square both sides:
[√(x + 7) - 2√(x)] * [√(x + 7) - 2√(x)] = 4
so then I FOIL the left side, resulting in:
(x+7) - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] + 4x
So then I thought to subtract (x+7) and (4x) to both sides
 - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] = 4 - 4x - x - 7 (I think I'm supposed to switch the sign, because I've
subtracted it and moved it to the opposite side, right?)
I think I'm correct up to this point, but now I have to square both sides again.
I think this left hand side could be re-written as:
 -2[√(x + 7) * - 2√(x)]
Is this right? I'm subtracting it from itself, a negative, which could simply multiplied by -2. Anyway, now I need to
square this again, so I assume the -2 becomes a 4 and I FOIL them separately?
FOILING the left side will get:
 [√(x + 7) * - 2√(x)] * [√(x + 7) * - 2√(x)]
Which, when FOILed, looks like
(x+7) - [√(x+7) * -2√(x)] - [ - 2√(x) * √(x + 7)] + 4x
It looks exactly the same as before!! I'm just really confused by this problem, and I have a couple more like it, so I
want to know if figuring this one out could help me solve the other ones.

I'm confused about FOILing the different sides, whether or not I can combine two square roots, and quite frankly, a lot
of other things.
One of the options on the test is 9, and I think this is the answer, because I've inserted it into the original
equation and it works, but I'm just confused about how to actually get 9 out of this..

Sorry for the long question. I hope it's not hard to understand. I'm just hoping someone can walk me through all
the steps of solving a problem like this so I can do it easily in the future.
*******************************
Here is my problem.
[SQRT(x + 7)] - 2[SQRT(x)] =-2      OR       √(x + 7) - 2√(x) = -2 
-------
Now, the steps I took to solve this problem are to first square both sides:
[√(x + 7) - 2√(x)] * [√(x + 7) - 2√(x)] = 4
so then I FOIL the left side, resulting in:
(x+7) - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] + 4x <=== This is where you went WRONG! When FOILed, this's
actually: , which results in:
                 x+%2B+7+-+2sqrt%28x%29sqrt%28x+%2B+7%29+-+2sqrt%28x%29sqrt%28x+%2B+7%29+%2B+4x+=+4. See?          
                                     x+%2B+7+-+4sqrt%28x%29sqrt%28x+%2B+7%29+%2B+4x+=+4                                                                                        
                                                    -+4sqrt%28x%29sqrt%28x+%2B+7%29+=+4+-+%28x+%2B+7%29+-+4x  So then I thought to subtract (x+7) and (4x) to both sides
                                                    -+4sqrt%28x%29sqrt%28x+%2B+7%29+=+4+-+x+-+7+-+4x%29  - [√(x + 7) * - 2√(x)] - [ - 2√(x) * √(x + 7)] = 4 - 4x - x - 7
                                                                                                     (I think I'm supposed to switch the sign, because I've 
                                                                                                     subtracted it and moved it to the opposite side, right?)
                                                  -+4sqrt%28x%5E2+%2B+7x%29+=+-+3+-+5x <=== This is exactly how you need to proceed!
                                           %28-+4sqrt%28x%5E2+%2B+7x%29%29%5E2+=+%28-+3+-+5x%29%5E2 ---- Squaring both sides
                                                 16%28x%5E2+%2B+7x%29+=+9+%2B+30x+%2B+25x%5E2
                                                  16x%5E2+%2B+112x+=+9+%2B+30x+%2B+25x%5E2
                       16x%5E2+%2B+112x+-+9+-+30x+-+25x%5E2+=+0
                                             -+9x%5E2+%2B+82x+-+9+=+0        OR       9x%5E2+-+82x+%2B+9+=+0
                                                                                            9x%5E2+-+81x+-+x+%2B+9+=+0
                                                                                       9x(x - 9) - 1(x - 9) = 0
                                                                                               (x - 9)(9x - 1) = 0
                                                                                                x - 9 = 0               OR      9x - 1 = 0
                                                                                                      x = 0 + 9        OR           9x = 1
                                                                                                      x = 9               OR              x = 1%2F9 (IGNORE)
1%2F9 proves to be EXTRANEOUS, so sole solution is: x = 9
To some though, it's easier to solve, if one of the left-side RADICALS is MOVED to the right, 1st. But then, this's subjective.


It looks exactly the same as before!! I'm just really confused by this problem, and I have a couple more like it, so I want
to know if figuring this one out could help me solve the other ones. Hopefully, the above will clear up SOME/ALL confusion!

I'm confused about FOILing the different sides, whether or not I can combine two square roots, and quite frankly, a lot of
other things.  Hopefully, the above will clear up SOME/ALL confusion!

One of the options on the test is 9, and I think this is the answer, because I've inserted it into the original equation
and it works, but I'm just confused about how to actually get 9 out of this.  Yes, the solution is indeed 9, as seen above! 

Sorry for the long question. I hope it's not hard to understand. I'm just hoping someone can walk me through all the steps
of solving a problem like this so I can do it easily in the future.  Hopefully, this author has assisted you in understanding this 
problem, so you can understand and obtain SOLUTIONS to similar problems, more easily, more efficiently, and without confusion!


Question 48636: Please help. These problems look so easy but I just can't seem to get the right answer.
First Problem:
1/√3 + 4/√27 - 2/√12

Second Problem:
2 cubed root of 27x + 2 cubed root 64x
I came up with 2 answers:
14 cubed root of x -OR- 2 cubed root of 91x
Which one is correct??

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
Please help. These problems look so easy but I just can't seem to get the right answer.
First Problem:
1/√3 + 4/√27 - 2/√12

Second Problem:
2 cubed root of 27x + 2 cubed root 64x
I came up with 2 answers:
14 cubed root of x -OR- 2 cubed root of 91x
Which one is correct??
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        In the post by @consc198, first solution is irrelevant,  while second solution is incorrect.

        For right solutions,  see what follows.


        F i r s t   p r o b l e m


It does not say what to do and what is required, but according to common sense, 
they want you to simplify and to reduce to simple canonic final form.


  1%2Fsqrt%283%29 + 4%2Fsqrt%2827%29 - 2%2Fsqrt%2812%29 = 

= 1%2Fsqrt%283%29 + 4%2F%283%2Asqrt%283%29%29 - 2%2F%282%2Asqrt%283%29%29 = 

= %281%2Fsqrt%283%29%29%2A%281+%2B+%284%2F3%29+-+1%29 = 4%2F%283%2Asqrt%283%29%29 = %284%2Asqrt%283%29%29%2F%283%2A3%29 = %284%2F9%29%2Asqrt%283%29.    ANSWER




        S e c o n d   p r o b l e m


  2%2Aroot%283%2C27x%29 + 2%2Aroot%283%2C64x%29 = 


       Notice that  27 = 3^3,  while  64 = 4^3.

       So, we continue the chain of equalities


= 2%2A3%2Aroot%283%2Cx%29+%2B+2%2A4%2Aroot%283%2Cx%29 = 6%2Aroot%283%2Cx%29 + 8%2Aroot%283%2Cx%29 = 14%2Aroot%283%2Cx%29.    the correct ANSWER

Solved.

As this person, @consc198, solves simple standard Math problems, he deserves to be excluded
from everywhere of an educational community, because he is not able to teach in a right way.




Question 1095268: how much xsquare -y square ? if x+y=66 xy=9

Found 4 solutions by KMST, ikleyn, greenestamps, MathTherapy:
Answer by KMST(5396) About Me  (Show Source):
You can put this solution on YOUR website!
ONE WAY:
Maybe we know about solving quadratic equations.
Sometimes we can solve a quadratic equation such as z%5E2%2Bbz%2Bc=0 by factoring if we find two integers p q such that p%2Bq=-b and pq=c .
Then those values for p and q are the solutions of the equation and the equation is really
z%5E2-%28p%2Bq%29z%2Bpq=%28z-p%29%28z-q%29=0 .
The solutions to a quadratic equation of the form z%5E2%2Bbz%2Bc=0 are always two numbers whose sum and product can be found are -b and c respectively.
Unfortunately, sometimes those numbers are irrational, or even imaginary, and then factoring is not an option.
Then, w must use algebra to "complete the square" and then solve, or apply the dreaded quadratic formula
that says the solutions to an equation of the form ax%5E2%2Bbx%2Bc=0 are given by
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
To avoid confusion, I used z for the variable instead of x , and I prefer equations where the leading coefficient is a=1 , so I wrote my equation as z%5E2%2Bbz%2Bc=0 .
I will make b=-%28p%2Bq%29=-66 and c=pq=9 to get z%5E2-66z%2B9=0.
The solutions will be the numbers p and q (or x and y ) that add up to x%2By=66 and whose product is x%2Ay=9 .
The quadratic formula tells me that the solution are given by

The solutions to the equation are %2833%2B6sqrt%2830%29%29 and %2833-12sqrt%2830%29%29
One is x and the other is y , but there is no way to guess which was intended to be x and which was intended to be y .
x%5E2 and y%5E2 are
and

The possible answers are
and

ANOTHER APPROACH:
We know that x%5E2-y%5E2=%28x%2By%29%28x-y%29 . If we only knew the value of %28x-y%29 we could easily find the value of x%5E2-y%5E2

We know that %28x-y%29%5E2=x%5E2%2By%5E2-2xy , but to calculate %28x-y%29%5E2 we would need the value of x%5E2%2By%5E2

We know that %28x%2By%29%5E2=x%5E2%2By%5E2%2B2xy and we know the values of %28x%2By%29=66=2%2A3%2A11 and xy=9=3%5E2
Substituting the known values we get 66%5E2=x%5E2%2By%5E2%2B2%2A9 --> 66%5E2=x%5E2%2By%5E2%2B18 --> x%5E2%2By%5E2=66%5E2-18=4338
Now we can use that value of to calculate the values of %28x-y%29%5E2 and %28x-y%29 , and from that find the value of x%5E2-y%5E2

%28x-y%29%5E2=x%5E2%2By%5E2-2xy=4338-2%2A9=4338-18=4320=30%2A144=30%2A12%5E2

Then %28x-y%29=%22+%22+%2B-+sqrt%2830-12%5E2%29=%22+%22+%2B-+12sqrt%2830%29

Multiplying times

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
How much x square -y square ? if x+y=66 xy=9
~~~~~~~~~~~~~~~~~~~~~~~


Tutor @greenestamps provided the answer  x%5E2-y%5E2 = 792sqrt%2830%29.

This answer is correct,  but incomplete.

This problem has a symmetry  (x.y) <---> (y,x).

In other words, values x and y can be rearranged.

It means that together with the answer  x%5E2-y%5E2 = 792sqrt%2830%29
 another answer  x%5E2-y%5E2 = -792sqrt%2830%29  is valid and is possible,  too.

So, the complete answer to the problem is THIS:

        - under given conditions,  x%5E2-y%5E2  may have two different values.
           One possible value is  792sqrt%2830%29;  another possible value is  -792sqrt%2830%29.



Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Given: x+y=66; xy=9

Find: x%5E2-y%5E2

x%5E2-y%5E2=%28x%2By%29%28x-y%29=66%28x-y%29



x%5E2-y%5E2=%2866%29%2812sqrt%2830%29%29=792sqrt%2830%29


Answer by MathTherapy(10858) About Me  (Show Source):

Question 316943: simplify: sqrt (121) / (9)
a) sqrt (11) / sqrt (3)
b) 3 sqrt (11)
c) 11/3
d) sqrt (11) / (3)

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
simplify: sqrt (121) / (9)

a) sqrt (11) / sqrt (3)
b) 3 sqrt (11)
c) 11/3
d) sqrt (11) / (3)
******************
The other person's answer, "...d because the sqrt of 121 is 11 and the sqrt of 9 is 3," is PARTIALLY WRONG!

If it's sqrt%28121%2F9%29, then it's matrix%281%2C3%2C+sqrt%28121%29%2Fsqrt%289%29+=+11%2F3%2C+%22%28CHOICE%22%2C+%22c%29%22%29

If it's sqrt%28121%29%2F9, then it's matrix%281%2C3%2C+11%2F9%2C+%22%28no%22%2C+%22match%29%22%29


Question 1030250: I need help simplifying this expression:
sqrt%2824%2B8sqrt%285%29%29
I removed the common factor out of the square root to obtain 2sqrt%286%2B2sqrt%285%29%29, but the answer key says it is 2%2B2sqrt%285%29.
How is it possible? Am I missing out on a rule here?

Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


Consider this patern:

%28sqrt%28a%29%2Bsqrt%28b%29%29%5E2=%28a%2Bb%29%2B2sqrt%28ab%29

Given an expression to be simplified in the form sqrt%28a%2Bb%2Asqrt%28c%29%29, if you can modify the expression so the coefficient "b" on the inside radical is 2, then the expression can be simplified using that pattern.

In your example, you can take "4" out of the coefficient "8", leaving the required coefficient "2"; the "4" goes back inside the radical as sqrt%2816%29:

sqrt%2824%2B8sqrt%285%29%29=sqrt%2824%2B2sqrt%2816%2A5%29%29=sqrt%2824%2B2sqrt%2880%29%29

The pattern now says you want to find integers a and b for which a+b=24 and ab=80. Those integers are 20 and 4, so




Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
I need help simplifying this expression:
sqrt%2824%2B8sqrt%285%29%29
I removed the common factor out of the square root to obtain 2sqrt%286%2B2sqrt%285%29%29, but the answer key says it is 2%2B2sqrt%285%29.
How is it possible? Am I missing out on a rule here?
*************************************
This happens to be one of those RARE cases when you can actually factor out a COMMON factor, which happens to be a
PERFECT SQUARE (in 4). This is what you did:
sqrt%2824+%2B+8sqrt%285%29%29 = sqrt%284%286+%2B+2sqrt%285%29%29%29 
                            sqrt%284%29+%2A+sqrt%286+%2B+2sqrt%285%29%29 ----- Applying sqrt%28m%2An%29 = sqrt%28m%29sqrt%28n%29         
                               2sqrt%286+%2B+2sqrt%285%29%29....At this point, you haven't fully simplified the
                                                       expression, and should've continued as follows:  
                            
                   2sqrt%285+%2B+1+%2B+2sqrt%285%2A1%29%29 ---- Changing 6 to 5 + 1, and 5 to 5*1  
                    2sqrt%285+%2B+1+%2B+2sqrt%285%29sqrt%281%29%29 ---- Applying sqrt%28m%2An%29 = sqrt%28m%29sqrt%28n%29         
   2sqrt%28%28sqrt%285%29%29%5E2+%2B+%28sqrt%281%29%29%5E2+%2B+2sqrt%285%29sqrt%281%29%29 --- Converting 
The above is in the form: %28a+%2B+b%29%5E2, with system%28matrix%282%2C3%2C+a%2C+being%2C+sqrt%285%29%2C+b%2C+being%2C+sqrt%281%29%29%29, and so:
2sqrt%28%28sqrt%285%29%29%5E2+%2B+%28sqrt%281%29%29%5E2+%2B+2sqrt%285%29sqrt%281%29%29 then becomes: 2sqrt%28%28sqrt%285%29+%2B+sqrt%281%29%29%5E2%29 
                                                                                2%28sqrt%285%29+%2B+sqrt%281%29%29 ----- Cancelling SQUARE and SQUARE ROOT
                                                                                2%28sqrt%285%29+%2B+1%29 = highlight%282sqrt%285%29+%2B+2%29
                                    This certainly matches the answer key: 2%2B2sqrt%285%29
=====
On the other hand, this author would've SIMPLIFIED the SURD from the onset, as follows:
                                 sqrt%2824+%2B+8sqrt%285%29%29
                       sqrt%2824+%2B+2%284%29sqrt%285%29%29 ----- Replacing 8 with factors, 2 & 4
                         sqrt%2824+%2B+2sqrt%2816%29sqrt%285%29%29 ----- Converting 4 to sqrt%2816%29
                              sqrt%2824+%2B+2sqrt%2880%29%29 ----- Applying sqrt%28m%29sqrt%28n%29 = sqrt%28mn%29%29
                        sqrt%2824+%2B+2sqrt%2820%2A4%29%29                
                    sqrt%2820+%2B+4+%2B+2sqrt%2820%29sqrt%284%29%29 ---- Changing 24 to 20 + 4, and applying sqrt%28m%2An%29 = sqrt%28m%29sqrt%28n%29           
  sqrt%28%28sqrt%2820%29%29%5E2+%2B+%28sqrt%284%29%29%5E2+%2B+2sqrt%2820%29sqrt%284%29%29 ---- Converting 
The above is in the form: %28a+%2B+b%29%5E2, with system%28matrix%282%2C3%2C+a%2C+being%2C+sqrt%2820%29%2C+b%2C+being%2C+sqrt%284%29%29%29, and so:
    sqrt%28%28sqrt%2820%29%29%5E2+%2B+%28sqrt%284%29%29%5E2+%2B+2sqrt%2820%29sqrt%284%29%29 then becomes:      sqrt%28%28sqrt%2820%29+%2B+sqrt%284%29%29%5E2%29
                                                                                                 sqrt%2820%29+%2B+sqrt%284%29 ----- Cancelling SQUARE and SQUARE ROOT
                                                                                                 highlight%282sqrt%285%29+%2B+2%29
                              This certainly matches the answer key: 2%2B2sqrt%285%29


Question 973976: what is (4- square root of 8) divided by (2+ square root of 8) simplified

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
what is (4- square root of 8) divided by (2+ square root of 8) simplified
*********************************************

According to the other person, "I put this into the calculator at Mathway.com, and it comes out as:
−4+3√2 (that's "three square root two")." 

Is this really "HELP?" Punching numbers on a calculator? Or, is this OUTRIGHT LAZINESS?

%284+-+sqrt%288%29%29%2F%282+%2B+sqrt%288%29%29
%284+-+sqrt%288%29%29%2F%282+%2B+sqrt%288%29%29 * %282+-+sqrt%288%29%29%2F%282+-+sqrt%288%29%29 ---- Multiplying NUMERATOR and DENOMINATOR by 2+-+sqrt%288%29, the DENOMINATOR'S conjugate
%284+-+sqrt%288%29%29%282+-+sqrt%288%29%29%2F%282+%2B+sqrt%288%29%29%2F%282+-+sqrt%288%29%29
 
        %288+-+6sqrt%288%29+%2B+8%29%2F%284+-+8%29
        %2816+-+12sqrt%282%29%29%2F%28-+4%29 ---- Converting 6sqrt%288%29 to 12sqrt%282%29
    %28-+4%28-+4+%2B+3sqrt%282%29%29%29%2F%28-+4%29 
    %28cross%28-+4%29%28-+4+%2B+3sqrt%282%29%29%29%2F%28cross%28-+4%29%29 = highlight%28-+4+%2B+3sqrt%282%29%29


Question 78896: please help me simplify this problem?
sqrt75a%5E2/sqrt5

Found 3 solutions by ikleyn, timofer, MathTherapy:
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
please help me simplify this problem?
sqrt%2875%2Aa%5E2%29%2Fsqrt%285%29
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


The right way to simplify is THIS

        sqrt%2875%2Aa%5E2%29%2Fsqrt%285%29 = sqrt%2815a%5E2%29 = abs%28a%29%2Asqrt%2815%29 = sqrt%2815%29%2Aabs%28a%29.         ANSWER


                Solved completely and correctly.


According to commonly accepted agreements about square root,
the expression  ' a%5E2 '  always goes from the square root as  |a|,  the absolute value of  'a'.

In this form,  this transformation works universally for positive and negative values of  ' a '.

The given problem is a standard  TRAP  to catch  highlight%28highlight%28millions%29%29  of those,  who are unfamiliar with this trap for advance.

My post shows and teaches a standard way to avoid this trap.


As tutor @MathTherapy likes to say,  " do not accept any other answer ".



Answer by timofer(159) About Me  (Show Source):
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
please help me simplify this problem?
sqrt75a%5E2/sqrt5
*************************************
The other person wrote, "Now, you need to see that 75=5%2A4%2A4, so you can also pull out a 4 from under the radical"
How does this make sense? In what universe is 75=5%2A4%2A4? As a result, his answer, 4a is obviosly WRONG!!

highlight%28sqrt%2875a%5E2%29%2Fsqrt%285%29%29 = sqrt%2875a%5E2%2F5%29 = sqrt%2815a%5E2%29 = highlight%28matrix%281%2C4%2C+a%2Asqrt%2815%29%2C%22%2C%22%2C+or%2C+sqrt%2815%29a%29%29


Question 669305: Please help me simplify this: sqrt(x^4y^7)
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me simplify this: sqrt(x^4y^7)
*************************************
The other person's answer, highlight_green%28x%5E2y%5E2sqrt%28y%5E3%29%29, is WRONG as he/she failed to SIMPLIFY this expression, completely. 

highlight%28sqrt%28x%5E4y%5E7%29%29 = sqrt%28%28x%5E2%29%5E2%28y%5E3%29%5E2y%29%29 = highlight%28x%5E2y%5E3sqrt%28y%29%29


Question 60849: solve for x by using the quadratic formula: 9x^2-6x+5=0
thanks so much

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
solve for x by using the quadratic formula: 9x^2-6x+5=0
thanks so much
~~~~~~~~~~~~~~~~~~~~~~~

        The solution by @jai_kos is incorrect.
        See my correct solution below.


9x^2-6x+5=0

It is the quadratic equation of the form.

ax^2+bx +c =0

Comparing the equations, we get
a = 9, b = -6 and c = 5

x = %28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

x = %28-%28-6%29+%2B-+sqrt%28+%28-6%29%5E2-4%2A9%2A5+%29%29%2F%282%2A9%29+

x = %286+%2B-+sqrt%28+36-180+%29%29%2F18+

x = %286+%2B-+sqrt%28-144%29%29%2F18+

x = %286+%2B-+12i%29%2F18+ = %281+%2B-+2i%29%2F3.

Thus the given equation has no real solutions.
It has two conjugate solutions in complex numbers.

Solved correctly.




Question 60848: help on this please,,thanks
x+1= sqrt x+7

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
help on this please,, thanks
x+1= sqrt(x+7)
~~~~~~~~~~~~~~~~~~~~~~~~


        The solution in the post by @jai_kos is incorrect.
        See my correct solution below.


x+1= sqrt(x+7)

Square on both sides, we get

(x+1)^2 = x +7
x^2 + 2x + 1 = x + 7

Reduce it to the standard form quadratic equation

x^2 + x - 6 = 0

Factor left side

(x+3)*(x-2) = 0

The roots of this equation are x = -3 and x = 2.
Direct check shows that x = 2 is the solution to the original equation, while x = -3 is an EXTRANEOUS solution.


ANSWER. The given equation has a unique solution in real numbers x = 2.


Solved correctly.




Question 681800: I'm trying to prove that
sqrt%284+-+2+%2A+sqrt%283%29%29+%2B+sqrt%2812+-+6+%2A+sqrt%283%29%29+=+2

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
I'm trying to prove that

sqrt%284+-+2+%2A+sqrt%283%29%29+%2B+sqrt%2812+-+6+%2A+sqrt%283%29%29+=+2            
*******************
This author's proof is different from the one the other person who responded, provided. His method
may not be what the person who needs help, is looking for. Then again, this hunch may be WRONG!

Having said that, let's focus on the L.H.S. because that's what's needed to be proven to equal 2.
                        sqrt%284+-+2sqrt%283%29%29 + sqrt%2812+-+6sqrt%283%29%29
                        sqrt%284+-+2sqrt%283%29%29 + sqrt%2812+-+2%283%29sqrt%283%29%29 ---- Changing 6 to 2(3)
                        sqrt%284+-+2sqrt%283%29%29 + sqrt%2812+-+2sqrt%289%29sqrt%283%29%29 ------- Changing 3 to sqrt%289%29 
               sqrt%283+%2B+1+-+2sqrt%283%29sqrt%281%29%29 + sqrt%289+%2B+3+-+2sqrt%289%29sqrt%283%29%29 ----- Changing 4 to 3 + 1, 3 to sqrt%283%29sqrt%281%29, and 12 to 9 + 3                 
sqrt%28%28sqrt%283%29%29%5E2+%2B+%28sqrt%281%29%29%5E2+-+2sqrt%283%29sqrt%281%29%29 + sqrt%28%28sqrt%289%29%29%5E2+%2B+%28sqrt%283%29%29%5E2+-+2sqrt%289%29sqrt%283%29%29 ------ Converting 
The above is in the form:
                          %28a%5B1%5D+-+b%5B1%5D%29%5E2  + %28a%5B2%5D+-+b%5B2%5D%29%5E2, with , and so:
sqrt%28%28sqrt%283%29%29%5E2+%2B+%28sqrt%281%29%29%5E2+-+2sqrt%283%29sqrt%281%29%29 + sqrt%28%28sqrt%289%29%29%5E2+%2B+%28sqrt%283%29%29%5E2+-+2sqrt%289%29sqrt%283%29%29 then becomes:
                   sqrt%28%28sqrt%283%29+-+sqrt%281%29%29%5E2%29 +  sqrt%28%28sqrt%289%29+-+sqrt%283%29%29%5E2%29
                         sqrt%283%29+-+sqrt%281%29    +    sqrt%289%29+-+sqrt%283%29 ----- Cancelling SQUARE and SQUARE ROOT
                             sqrt%283%29+-+1 + 3+-+sqrt%283%29+=+highlight%282%29%29QED!


Question 1107093: One pair of integers (x.y) solves +sqrt%28+4%2Asqrt%28+7+%29%2B11+%29=y%2Bsqrt%28+x+%29+, find x*y.
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
One pair of integers (x.y) solves +sqrt%28+4%2Asqrt%28+7+%29%2B11+%29=y%2Bsqrt%28+x+%29+, find x*y.
***************************************************************
                              sqrt%284sqrt%287%29+%2B+11%29%29
                              sqrt%2811+%2B+4sqrt%287%29%29
                    sqrt%2811+%2B+2%282%29sqrt%287%29%29 ----- Replacing 4 with its PRIME factors, 2 & 2
                         sqrt%2811+%2B+2sqrt%284%29sqrt%287%29%29 ----- Converting 2 to sqrt%284%29                 
                      sqrt%287+%2B+4+%2B+2sqrt%287%29sqrt%284%29%29 ---- Changing 11 to 7 + 4             
     sqrt%28%28sqrt%287%29%29%5E2+%2B+%28sqrt%284%29%29%5E2+%2B+2sqrt%287%29sqrt%284%29%29 ---- Converting 
The above is in the form: %28a+%2B+b%29%5E2, with system%28matrix%282%2C3%2C+a%2C+being%2C+sqrt%287%29%2C+b%2C+being%2C+sqrt%284%29%29%29, and so:
sqrt%28%28sqrt%287%29%29%5E2+%2B+%28sqrt%284%29%29%5E2+%2B+2sqrt%287%29sqrt%284%29%29 then becomes: sqrt%28%28sqrt%287%29+%2B+sqrt%284%29%29%5E2%29 
                                                                                sqrt%287%29+%2B+sqrt%284%29 ----- Cancelling SQUARE and SQUARE ROOT
                                                                                sqrt%287%29+%2B+2

√(4√(7+11)) = sqrt%287%29+%2B+2, is in the form: y + √x, with y = 2, and x = 7. Therefore, x*y = 7(2) = 14. 


Question 1124474: How would you simplify the equation:
Sqrt(17-12sqrt2) in the form
(a+bsqrt2)
Thank you.

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
How would you simplify the equation:
Sqrt(17-12sqrt2) in the form
(a+bsqrt2)
Thank you. 
**********************************
Not an equation, but an expression, instead!!
                           sqrt%2817+-+12sqrt%282%29%29
                    sqrt%2817+-+2%286%29sqrt%282%29%29 ----- Changing 12 to 2*6
                     sqrt%2817+-+2sqrt%2836%29sqrt%282%29%29 ------ Converting 6 to sqrt%2836%29
                         sqrt%2817+-+2sqrt%2872%29%29 ------ Applying sqrt%28m%29sqrt%28n%29+=+sqrt%28mn%29
                  sqrt%289+%2B+8+-+2%289%2A8%29%29 ----- Changing 17 to 9 + 8, and 72 (in sqrt%2872%29) to 9*8
                    sqrt%289+%2B+8+-+2sqrt%289%29sqrt%288%29%29 ------ Applying sqrt%28m%2An%29+=+sqrt%28m%29sqrt%28n%29
  sqrt%28%28sqrt%289%29%29%5E2+%2B+%28sqrt%288%29%29%5E2+-+2sqrt%289%29sqrt%288%29%29 ------ Converting 
The above is in the form: %28a+-+b%29%5E2, with system%28matrix%282%2C3%2C+a%2C+being%2C+sqrt%289%29%2C+b%2C+being%2C+sqrt%288%29%29%29, and so:
sqrt%28%28sqrt%289%29%29%5E2+%2B+%28sqrt%288%29%29%5E2+-+2sqrt%289%29sqrt%288%29%29 then becomes: sqrt%28%28sqrt%289%29+-+sqrt%288%29%29%5E2%29 
                                                                                sqrt%289%29+-+sqrt%288%29 ----- Cancelling SQUARE and SQUARE ROOT
                                                                                highlight%283+-+2sqrt%282%29%29


Question 263259: solve: x^5/6 + x^2/3-2x^1/2 =0
thanks!

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
solve: x^5/6 + x^2/3 - 2x^1/2 = 0
~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution in the post by @mananth is FATALLY incorrect.

        His first step is to raise all three addends in the left side of the equation to degree 6.
        But this step is not an equivalent transformation, so, starting from this point,
        his solution is inadequate.

        See my correct solution below.


The original equation is 

    x^5/6 + x^2/3 - 2x^1/2 = 0.      (1)


It is the same as

    x^5/6 + x^4/6 - 2x^3/6 = 0.      (2)


Factor left side, taking  x^3/6  out the parentheses as a common factor

    x^3/6*(x^2/6 + x^1/6 - 2) = 0.   (3)


One root is  x = 0,  generated by the common factor  x^3/6.


To find the roots generated by the expression in parentheses, introduce new variable  t = x^1/6.


Then the expression in parentheses is 

    t^2 + r - 2.


Its roots are  t%5B1%2C2%5D = %28-1+%2B-+sqrt%281+-+4%2A1%2A%28-2%29%29%29%2F2 = %28-1+%2B-+sqrt%281%2B8%29%29%2F2 = %28-1+%2B+sqrt%289%29%29%2F2 = %28-1+%2B-+3%29%2F2.


So, the roots are t = -2 and  t = 1.   


Since t = x^1/6,  we accept the positive root  t = 1  and  reject the negative root t = -2.


Thus, finally we have the solutions to the original equation  x = 0  and  x = 1.


ANSWER.  The original equation has two real solutions  x = 0  and  x = 1.

Solved correctly.




Question 1094302: Solving by substitution with u
+x%5E2-3x-sqrt%28x%5E2-3x%29=2+
+u+=+sqrt%28x%5E2-3x%29+
+u%5E2+=+sqrt%28x%5E2-3x%29+=+x%5E2-3x+
creates quadratic = +u%5E2-u+=+2+
+u%5E2-u-2=0+
+%28u-2%29%28u%2B1%29+
u = 2,-1
going back and using those two outputs in +sqrt%28x%5E2-3x%29+
+sqrt%28%282%29%5E2-3%282%29%29+ and +sqrt%28%28-1%29%5E2-3%28-1%29%29+
+sqrt%284-6%29+ and +sqrt%281%2B3%29+
Finally, I get +sqrt%28-2%29+ and +sqrt%284%29+
But -2 can't be in a radical and would 4 turn into 2,-2 or just 2?

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Solving by substitution with u
+x%5E2-3x-sqrt%28x%5E2-3x%29=2+

+u+=+sqrt%28x%5E2-3x%29+
+u%5E2+=+sqrt%28x%5E2-3x%29+=+x%5E2-3x+
creates quadratic = +u%5E2-u+=+2+
+u%5E2-u-2=0+
+%28u-2%29%28u%2B1%29+
u = 2,-1
going back and using those two outputs in +sqrt%28x%5E2-3x%29+
+sqrt%28%282%29%5E2-3%282%29%29+ and +sqrt%28%28-1%29%5E2-3%28-1%29%29+
+sqrt%284-6%29+ and +sqrt%281%2B3%29+
Finally, I get +sqrt%28-2%29+ and +sqrt%284%29+

But -2 can't be in a radical and would 4 turn into 2,-2 or just 2? 
******************************************************************

u = 2,-1 <=== This is okay!
"going back and using those two outputs in +sqrt%28x%5E2-3x%29+    +sqrt%28%282%29%5E2-3%282%29%29+ and +sqrt%28%28-1%29%5E2-3%28-1%29%29+ <=== Here's where I guess 
                                                                                                                you got confused, and substituted 2 and - 1 for x in sqrt%28x%5E2-3x%29

But, u = 2, as you mentioned above, NOT x = 2. And, because you'd substituted u for sqrt%28x%5E2+-+3x%29 earlier, at this juncture, you
need to BACK-SUBSTITUTE the value of u to get:
                      2+=+sqrt%28x%5E2+-+3x%29 
                   2%5E2+=+%28sqrt%28x%5E2+-+3x%29%29%5E2 ---- Squaring both sides
                    4+=+x%5E2+-+3x
     x%5E2+-+3x+-+4+=+0
(x - 4)(x + 1) = 0
  x - 4 = 0          OR        x + 1 = 0
       x = 4           OR              x = - 1 

Now, you have 2 values for x that you can CHECK to ensure that they're VALID and NOT EXTRANEOUS. 

Also, u = - 1, as you mentioned above. And, because you'd substituted u for sqrt%28x%5E2+-+3x%29 earlier, at this juncture, you need 
to BACK-SUBSTITUTE the value of u to get: -+1+=+sqrt%28x%5E2+-+3x%29 
Seeing that the square root of ANY expression is positive (> 0), it's obvious that u = - 1 is an EXTRANEOUS value. As such,
x = 4, or x = - 1 (see above). 


Question 569284: solve the following radical: +sqrt+%28+3x+%29%2B+sqrt+%28+2x-1+%29+=+5+%2F+sqrt+%28+2x-1+%29+
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
solve the following radical: +sqrt+%28+3x+%29%2B+sqrt+%28+2x-1+%29+=+5+%2F+sqrt+%28+2x-1+%29+
**********************************************<

Just like "Rational-functions/875226", - 12 is NOT a solution. It's an 
EXTRANEOUS root. Only VALID and ACCEPTABLE answer: highlight%28x+=+3%2F2%29


Question 864099: Can you show me how to Rationalize the denominator and simply the equation
6 / the square root of 2 - the square root of 3.

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Can you show me how to Rationalize the denominator and simply the equation 
6 / the square root of 2 - the square root of 3.
*********************************************************************
%28highlight%28sqrt%286%29%2F%28sqrt%282%29+-+sqrt%283%29%29%29%29%28%28sqrt%282%29+%2B+sqrt%283%29%29%2F%28sqrt%282%29+%2B+sqrt%283%29%29%29 =  = %28sqrt%2812%29+%2B+sqrt%2818%29%29%2F%282+-+3%29 = 


Question 864097: Can you show me how to Rationalize the denominator and simply the equation the square root of 6 / the square root of 5 - the square root of 3.
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Can you show me how to Rationalize the denominator and simply the equation
the square root of 6 / the square root of 5 - the square root of 3.
*********************************************************************
%28highlight%28sqrt%286%29%2F%28sqrt%285%29+-+sqrt%283%29%29%29%29%28%28sqrt%285%29+%2B+sqrt%283%29%29%2F%28sqrt%285%29+%2B+sqrt%283%29%29%29 =  = %28sqrt%2830%29+%2B+sqrt%2818%29%29%2F%285+-+3%29 = highlight%28%28sqrt%2830%29+%2B+3sqrt%282%29%29%2F2%29


Question 707329: How would you solve this equation?
+sqrt+%28+2x+%29+=+sqrt+%28+3x%2B12+%29-2+

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
How would you solve this equation?

+sqrt+%28+2x+%29+=+sqrt+%28+3x%2B12+%29-2+
**************************************
The other person's response, that "You get -10+6i or -10-6i (complex results)", is WRONG11

sqrt%282x%29+=+sqrt%283x%2B12%29+-+2
As the smaller radicand, 2x MUST be greater than or equal to 0, we get: 2x+%3E=+0. So, x+%3E=+0.  
This gives us:      sqrt%282x%29+=+sqrt%283x%2B12%29+-+2, with x+%3E=+0
                         %28sqrt%282x%29%29%5E2+=+%28sqrt%283x+%2B+12%29+-+2%29%5E2 ---- Squaring both sides
                               2x+=+3x+%2B+12+-+4sqrt%283x+%2B+12%29+%2B+4
                    4sqrt%283x+%2B+12%29+=+3x+%2B+12+%2B+4+-+2x
                    4sqrt%283x+%2B+12%29+=+x+%2B+16
                  16%283x+%2B+12%29+=+x%5E2+%2B+32x+%2B+256 ----- Squaring both sides
                     48x+%2B+192+=+x%5E2+%2B+32x+%2B+256
x%5E2+%2B+32x+%2B+256+-+48x+-+192+=+0
                 x%5E2+-+16x+%2B+64+=+0
                        %28x+-+8%29%5E2+=+0
                          x - 8 = 0.
                               highlight%28x+=+8%29 <==== VALID, and ACCEPTABLE solution for x, in THIS case!!


Question 144143: Hello. Please help me solve this problem: I need to find the simplest radical form and the approximate answer of sqrt+%285%29-1%2Fx+=+sqrt+%285%29%2F2.
2%28sqrt%285%29-2%2Fx=sqrt%285%29
sqrt%285%29-2%2Fx=0
sqrt%285%29=2%2Fx
x%28sqrt%285%29%29=2
x=2%2Fsqrt%285%29
x = 0.8944

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Hello. Please help me solve this problem: 
I need to find the  simplest radical form and the approximate answer of sqrt+%285%29-1%2Fx+=+sqrt+%285%29%2F2.
2%28sqrt%285%29-2%2Fx=sqrt%285%29
sqrt%285%29-2%2Fx=0
sqrt%285%29=2%2Fx
x%28sqrt%285%29%29=2
x=2%2Fsqrt%285%29
x = 0.8944
*****************
         sqrt%285%29+-+1%2Fx+=+sqrt%285%29%2F2
    2x%2Asqrt%285%29+-+2+=+x%2Asqrt%285%29 ---- Multiplying by LCD, 2x
2x%2Asqrt%285%29+-+x%2Asqrt%285%29+=+2 
          x%2Asqrt%285%29+=+2
      %28x%2Asqrt%285%29%29%5E2+=+2%5E2 ----- Squaring each side
       x%5E2%285%29+=+4
           x%5E2+=+4%2F5
            

However, the negative x-value, -2sqrt%285%29%2F5 is EXTRANEOUS, therefore leaving the sole VALID x-value, highlight%282sqrt%285%29%2F5%29. 

Great job!! You got up to this point: x=2%2Fsqrt%285%29, but needed to go a little further, by RATIONALIZING the
DENOMINATOR, as demonstrated above. Your decimal approximation, x = 0.8944, is also CORRECT.
I don't know why you sought help. You didn't need it!

Again, great job!!


Question 798582: Simplify if necessary. Then rationalize the denominator.
sqrt%282%2F3%29%2Asqrt%286%2F22%29

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Simplify if necessary. Then rationalize the denominator.

sqrt%282%2F3%29%2Asqrt%286%2F22%29
**************************

CORRECT answer: highlight%282sqrt%2811%29%2F11%29, or highlight%28%282%2F11%29sqrt%2811%29%29, and NOT sqrt%282%2F11%29 as the other person states.


Question 459722: Simplify each expression by rationalizing the denominator.
3/sqrt(7)
2sqrt(2)/sqrt(5)
3sqrt(2)/sqrt(6)
2sqrt(5)/sqrt(12)

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Simplify each expression by rationalizing the denominator.
3/sqrt(7)
2sqrt(2)/sqrt(5)
3sqrt(2)/sqrt(6)
2sqrt(5)/sqrt(12)
******************

This problem "asks" to SIMPLIFY by "rationalizing the denominator," NOT providing a calculated value, as the other person did!

      , or highlight%28%283%2F7%29sqrt%287%29%29

, or highlight%28%282%2F5%29sqrt%2810%29%29



, or highlight%28%281%2F3%29sqrt%2815%29%29


Question 904562: sqrt%284x%2B1%29-sqrt%282x%2B4%29=1
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%284x%2B1%29-sqrt%282x%2B4%29=1
*****************************
It is UNBELIEVABLE that the person who responded to this problem applied this RIDICULOUS approach. 

The person got up to x%5E2+-+6x+=+0, and UNBELIEVABLY, HE chose to, "complete the squares........"
What would behoove someone to suggest such a thing to a person who asks for help with a math problem?
Now, isn't x%5E2+-+6x+=+0 SIMPLY x(x - 6) = 0, which results in values of 0 and 6 for x? Why would someone
go through all that trouble to COMPLETE the SQUARE here? It's the same as using the quadratic equation formula
to solve it, which is EXTREMELY RIDICULOUS and TOTALLY UNNECESSARY.....a TOTAL waste of time, in this author's
opinion. Some people are probably more eccentric than is realized!!


Question 296793: Please help me solve this equation
4%2Bsqrt%2810-x%29=6%2Bsqrt%284-x%29

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve this equation
4%2Bsqrt%2810-x%29=6%2Bsqrt%284-x%29
******************************
This is UTTERLY CRAZY!! Who ever heard of this: 4%2Bsqrt%2810-x%29=6%2Bsqrt%284-x%29 square everything
4%5E2%2B%28sqrt%2810-x%29%29%5E2=6%5E2%2B%28sqrt%284-x%29%29%5E2 <=== Who in their right mind would square both sides of a square root problem like this? Why do
these people respond in this manner? This "HELP" is NO HELP at all!! Just leave the problem for someone who can handle it!
BTW, the person claimed this to be an UNDEFINED equation. In other words, there are no solutions! Really!! Ha-ha-ha! 

 4+%2B+sqrt%2810+-+x%29+=+6+%2B+sqrt%284+-+x%29
    sqrt%2810+-+x%29+=+2+%2B+sqrt%284+-+x%29
 %28sqrt%2810+-+x%29%29%5E2+=+%282+%2B+sqrt%284+-+x%29%29%5E2 --- Squaring BOTH sides, the CORRECT way
      10+-+x+=+4+%2B+4sqrt%284+-+x%29+%2B+4+-+x
10+-+8+-+x+%2B+x+=+4sqrt%284+-+x%29
         2+=+4sqrt%284+-+x%29
         1+=+2sqrt%284+-+x%29 ---- Dividing by GCF, 2
         1 = 4(4 - x) --- Squaring each side
         1 = 16 - 4x
    1 - 16 = - 4x
      - 15 = - 4x
    %28-+15%29%2F%28-+4%29+=+highlight%2815%2F4+=+x%29

This x-value, 15%2F4 is VALID!!


Question 37745: Hi, I'm having trouble with this problem. Do you think you could help me?
+sqrt+6x%5E3++%2A++sqrt+150x%5E2+

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, I'm having trouble with this problem. Do you think you could help me? 
+sqrt+6x%5E3++%2A++sqrt+150x%5E2+
***********************************************************************
Correct answer: highlight%2830x%5E2sqrt%28x%29%29, and NOT x%5E2sqrt%2830x%29 as the other person states!


Question 729072: square root of x+7=square root of 2x-3 (+2) not under the square root)
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
square root of x+7=square root of 2x-3    (+2) not under the square root)
*************************************************************************
I don't know what "X=507+338√2 ≈985.0042 from @lynnlo(4176) is all about.

square root of x+7=square root of 2x-3    (+2) not under the square root)
              sqrt%28x+%2B+7%29+=+sqrt%282x+-+3%29+%2B+2
            %28sqrt%28x+%2B+7%29%29%5E2+=+%28sqrt%282x+-+3%29+%2B+2%29%5E2 --- Squaring both sides
                x+%2B+7+=+2x+-+3+%2B+4sqrt%282x+-+3%29+%2B+4 
                x+%2B+7+=+2x+%2B+4sqrt%282x+-+3%29+%2B+1
         x+%2B+7+-+2x+-+1+=+4sqrt%282x+-+3%29
              -+x+%2B+6+=+4sqrt%282x+-+3%29
            %28-+x+%2B+6%29%5E2+=+%284sqrt%282x+-+3%29%29%5E2 ----- Squaring both sides
         x%5E2+-+12x+%2B+36+=+16%282x+-+3%29
         x%5E2+-+12x+%2B+36+=+32x+-+48
x%5E2+-+12x+%2B+36+-+32x+%2B+48+=+0
         x%5E2+-+44x+%2B+84+=+0
     (x - 42)(x - 2) = 0
      x - 42 = 0      OR     x - 2 = 0
           x = 42     OR         x = 2

However, x = 42 is an EXTRANEOUS solution, so ONLY solution is: x = 2


Question 519668: how do you solve the following:
sqrt%28x%2B4%29 + sqrt%282x-1%29 = sqrt%287x%2B1%29

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
how do you solve the following:

sqrt%28x%2B4%29 + sqrt%282x-1%29 = sqrt%287x%2B1%29
******************************
One of the persons who responded states that x+=+-+1%2F2, but this is WRONG!! x CANNOT equal -+1%2F2.

Of the 3 RADICAL expressions, 2x - 1 is the SMALLEST. So, 2x+-+1+%3E=+0, and 2x+%3E=+1 ===> x+%3E=+1%2F2.
  So, we now get: sqrt%28x+%2B+4%29 + sqrt%282x+-+1%29 = sqrt%287x+%2B+1%29, with x+%3E=+1%2F2
                  sqrt%28x+%2B+4%29 + sqrt%282x+-+1%29 = sqrt%287x+%2B+1%29
                  %28sqrt%28x+%2B+4%29+%2B+sqrt%282x+-+1%29%29%5E2 = %28sqrt%287x+%2B+1%29%29%5E2 ----- Squaring both sides
 = 7x + 1
     x+%2B+4+%2B+2sqrt%28%28x+%2B+4%29%282x+-+1%29%29+%2B+2x+-+1 = 7x + 1
              3x+%2B+3+%2B+2sqrt%282x%5E2+%2B+7x+-+4%29 = 7x + 1
                     2sqrt%282x%5E2+%2B+7x+-+4%29 = 7x + 1 - 3x - 3
                     2sqrt%282x%5E2+%2B+7x+-+4%29 = 4x - 2
                  %282sqrt%282x%5E2+%2B+7x+-+4%29%29%5E2 = %284x+-+2%29%5E2 ------ Squaring both sides
                       4%282x%5E2+%2B+7x+-+4%29+=+16x%5E2+-+16x+%2B+4
                         8x%5E2+%2B+28x+-+16+=+16x%5E2+-+16x+%2B+4
                                    0+=+16x%5E2+-+16x+%2B+4+-+8x%5E2+-+28x+%2B+16
                                    0+=+8x%5E2+-+44x+%2B+20
                                    0+=+4%282x%5E2+-+11x+%2B+5%29
                                    0+=+2x%5E2+-+11x+%2B+5
                                    0+=+2x%5E2+-+10x+-+x+%2B+5
                                    0+=+2x%28x+-+5%29+-+1%28x+-+5%29
                                    0 = (2x - 1)(x - 5) 
                                    0 = 2x - 1      OR     0 = x - 5
                                    1 = 2x          OR    5 = x
                                   highlight%281%2F2+=+x%29
As x+=+1%2F2+%3E=+1%2F2 and x+=+5+%3E=+1%2F2, both solutions for x are VALID! 


Question 701399: simplify +sqrt%2845%29-10%2Fsqrt%285%29+
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
simplify +sqrt%2845%29-10%2Fsqrt%285%29+
====================================
The other person's answer, 15, is WRONG!!

sqrt%2845%29+-+10%2Fsqrt%285%29 = %28sqrt%2845%29%2Asqrt%285%29%29%2Fsqrt%285%29+-+10%2Fsqrt%285%29 = %28sqrt%289+%2A+5%29sqrt%285%29+-+10%29%2Fsqrt%285%29 = %28sqrt%289%29sqrt%285%29sqrt%285%29+-+10%29%2Fsqrt%285%29 = %283%285%29+-+10%29%2Fsqrt%285%29 = %2815+-+10%29%2Fsqrt%285%29 = 5%2Fsqrt%285%29 = %285%2Fsqrt%285%29%29%28sqrt%285%29%2Fsqrt%285%29%29 = 5sqrt%285%29%2F5 = cross%285%29sqrt%285%29%2Fcross%285%29 = highlight%28sqrt%285%29%29


Question 993742: Hello. I am having an issue solving this problem and I look to you for help. My problem is:
6sqrt%2812%29-8sqrt%2850%29+9sqrt%2872%29. By using prime factorization, I know that 12 = 2^2*3
50 = 5^2*2
72 = 3^2*2^3
Now, I plug these terms back in and I get:
6%2A2sqrt%283%29-8%2A5sqrt%282%29+9%2A3%2A2sqrt%282%29 If I continue, I get:
12sqrt%283%29-40sqrt%282%29+54sqrt%282%29
The answer key shows that the answer should be
57sqrt%283%29-40sqrt%282%29 Can you see where I went wrong? I'm not coming up with the right answer.

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Hello. I am having an issue solving this problem and I look to you for help. My problem is:

6sqrt%2812%29-8sqrt%2850%29+9sqrt%2872%29. By using prime factorization, I know that 12 = 2^2*3
            50 = 5^2*2
            72 = 3^2*2^3
Now, I plug these terms back in and I get:
6%2A2sqrt%283%29-8%2A5sqrt%282%29+9%2A3%2A2sqrt%282%29 If I continue, I get:

12sqrt%283%29-40sqrt%282%29+54sqrt%282%29
The answer key shows that the answer should be
 57sqrt%283%29-40sqrt%282%29 Can you see where I went wrong? I'm not coming up with the right answer. 
=========================
57sqrt%283%29-40sqrt%282%29 can NEVER be the simplified answer here. This answer must be for a different problem. 
Correct answer: highlight%2812sqrt%283%29+%2B+14sqrt%282%29%29
=====================================================
Hello. I am having an issue solving this problem and I look to you for help. My problem is:

6sqrt%2812%29-8sqrt%2850%29+9sqrt%2872%29. By using prime factorization, I know that 12 = 2^2*3
            50 = 5^2*2
            72 = 3^2*2^3 
Now, I plug these terms back in and I get:
6%2A2sqrt%283%29-8%2A5sqrt%282%29+9%2A3%2A2sqrt%282%29    
6%2A2sqrt%283%29-8%2A5sqrt%282%29+9%2A3%2A2sqrt%282%29 If I continue, I get: 

  12sqrt%283%29-40sqrt%282%29+54sqrt%282%29
= 12sqrt%283%29+14sqrt%282%29<==== Should be THIS, your FINAL answer!!

You ALREADY have the correct answer, as you can see.


Question 1179144: For what values of c will x² +28x + c = 0 have no real solutions? Explain
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
For what values of c will x^2 +28x + c = 0 have no real solutions? Explain
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


The given quadratic equation has no real solutions if and only if its discriminant is a negative value.


The discriminant is  d = 28^2 - 4c = 784 - 4c.


The condition of negativity the discriminant is

    784 - 4c < 0,   or   4c > 784,   c > 784/4 = 196.


ANSWER.  The given equation has no real solutions at c > 196.

Solved.




Question 787079: Simplify the expression. Assume that all variables are positive
√x/5*√x/20

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Simplify the expression.  Assume that all variables are positive

√x/5*√x/20

If %28sqrt%28x%29%2F5%29+%2A+%28sqrt%28x%29%2F20%29, then the answer is NOT 10x, as stated by the other person who responded.
%28sqrt%28x%29%2F5%29+%2A+%28sqrt%28x%29%2F20%29 = %28sqrt%28x%29+%2A+sqrt%28x%29%29%2F%285+%2A+20%29 = sqrt%28x+%2A+x%29%2F100 = sqrt%28x%5E2%29%2F100%29 = highlight%28x%2F100%29


Question 99475: Perform the indicated division. Rationalize the denominator if necessary. Then simplify each radical expression
-9-sqrt(108) dived by 3

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Perform the indicated division. Rationalize the denominator if necessary. Then simplify each radical expression

-9-sqrt(108) dived by 3

The answer is highlight%28-+3+-+2sqrt%283%29%29%29 (see Question # 83444), and NOT -+5sqrt%283%29 as the other person states.

%28-9-6sqrt%283%29%29%2F3+%3C%3E+%28-15sqrt%283%29%29%2F3, as he/she suggests! I wonder where these people learned their mathematics!


Question 293344: how would you solve this proof?
+%28sqrt%286%29-sqrt%282%29%29%2F4+=+sqrt%282-sqrt%283%29%29%2F2+

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
how would you solve this proof?
+%28sqrt%286%29-sqrt%282%29%29%2F4+=+sqrt%282-sqrt%283%29%29%2F2+

This is indeed a NESTED SURD. From the other person's response, let me pick up from this point:
sqrt%288-4%2Asqrt%283%29%29+=+sqrt%286%29-sqrt%282%29+ Start with the given equation.

Let's focus on the L.H.S. of the equation.
sqrt%288+-+4sqrt%283%29%29
sqrt%288+-+2%2A2sqrt%283%29%29
sqrt%288+-+2sqrt%284%29sqrt%283%29%29 --- Converting 2 to sqrt%284%29
sqrt%288+-+2sqrt%284+%2A+3%29%29
sqrt%288+-+2sqrt%2812%29%29
sqrt%286+%2B+2+-+2sqrt%286+%2A+2%29%29 ---- Changing 8 to 6 + 2, and 12 to 6 * 2
sqrt%286+%2B+2+-+2sqrt%286%29sqrt%282%29%29
sqrt%28%28sqrt%286%29%29%5E2+%2B+%28sqrt%282%29%29%5E2+-+2sqrt%286%29sqrt%282%29%29 --- Converting 6 to %28sqrt%286%29%29%5E2 and 2 to %28sqrt%282%29%29%5E2
The above is now in the form: %28a+-+b%29%5E2, with a being sqrt%286%29, and b being sqrt%282%29
So, sqrt%28%28sqrt%286%29%29%5E2+%2B+%28sqrt%282%29%29%5E2+-+2sqrt%286%29sqrt%282%29%29 now becomes: sqrt%28%28sqrt%286%29+-+sqrt%282%29%29%29%5E2 
                                              sqrt%286%29+-+sqrt%282%29 ---- Cancelling square and sqrt
                                              highlight%28sqrt%286%29+-+sqrt%282%29%29 = highlight%28sqrt%286%29+-+sqrt%282%29%29
As seen, L.H.S. = R.H.S.


Question 485164: Write the following expression as a radical and simplify if possible.
(-64)^4/3

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Write the following expression as a radical and simplify if possible.
(-64)^4/3

The other person's answer, 81, is WRONG!!
highlight%28matrix%282%2C1%2C+%22+%22%2C+%28-+64%29%5E%284%2F3%29%29%29 = %28root%283%2C+-+64%29%29%5E4 = %28-+4%29%5E4+=+highlight%28256%29


Question 732218: how do you do √50x√18?
Found 2 solutions by MathTherapy, ikleyn:
Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
how do you do √50x√18?

√50 x √18 = 5√2 x 3√2 = 5(3)(√2 x √2) = 15√2 x √2 = 15√4 = 15(2) = 30

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
how do you do √50x√18?
~~~~~~~~~~~~~~~~~~~~~~

I treat the square roots as having positive values and 'x' as the multiplication sign.


It is what should be told in the problem's description, but since it is omitted,
I insert it to make your request sensible.


Then  sqrt%2850%29 * sqrt%2818%29 = sqrt%2850%2A18%29 = sqrt%28900%29 = 30.    ANSWER

Solved.




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