Questions on Algebra: Square root, cubic root, N-th root answered by real tutors!

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Question 1073123: A school buys 200 calculators at a special price. x of them were scientific calculators at R89 a piece and
the rest were financial calculators at R599 a piece. An expression, in terms of x, for the total number of
financial calculators bought is ...

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
A school buys 200 calculators at a special price. x of them were scientific calculators at R89 a piece and
the rest were financial calculators at R599 a piece. An expression, in terms of x, for the total number of
financial calculators bought is ...
****************************************
Total number of calculators purchased: 200
Number of scientific calculators purchased: x

Expression, in terms of x, for the total number of financial calculators purchased: 200 - x


Question 277236: What is the square root of 9y to the 6th power?
Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.
What is the square root of 9y to the 6th power?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


@mananth in his post treats it as   sqrt%289y%5E6%29,  and gives the answer   3y%5E3.

@dabanfield in his post treats it as   sqrt%28%289y%29%5E6%29,  and gives the answer   729y%5E3.


*******************************************

                They both are incorrect.

********************************************


If to treat it as   sqrt%289y%5E6%29,  then the correct answer is   3%2Aabs%28y%29%5E3  with the use of the absolute value sign.

If to treat it as   sqrt%28%289y%29%5E6%29,  then the correct answer is   729%2Aabs%28y%29%5E3  with the use of the absolute value sign.


The error of the two other persons is that value  ' y '  in the input formula can be negative.

Then both other persons return the negative value of the square root, which contradicts
to the common agreement about the square root values.

My formula works universally for positive and negative values of  ' y ' - - -  it is why it is preferable
and why it is the uniquely correct form.


Solved.


The problems of this kind are a standard TRAP to catch unprepared/undertrained novices.




Question 268365: i need help finding x in the equation: 2(squareroot x) + 3 = x
Found 4 solutions by MathTherapy, josgarithmetic, timofer, ikleyn:
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
i need help finding x in the equation: 2(squareroot x) + 3 = x
**************************************************************

The answer to this problem by one of the "respondents," timofer(153) AKA josgarithmetic(39765) is WRONG! See correct answer by @IKLEYN.

All other answers are BOGUS/FAKES, and should be REJECTED/IGNORED!

His answer: "(x=5%2B-+sqrt%2834%29-------------check if either or both will work...." 
BEWARE! You'll be checking from now to eternity and NEITHER would ever work!!

Answer by josgarithmetic(39792) About Me  (Show Source):
You can put this solution on YOUR website!
2%2Asqrt%28x%29%2B3=x

2%2Asqrt%28x%29=x-3
4x=x%5E2-6x%2B9
x%5E2-10x%2B9=0
discriminant 100-4%2A9=100-36=64
x=10%2B-+sqrt%2864%29%29%2F2
x=%2810%2B-+8%29%2F2
x=5%2B-+4
x=1 or x=9 -------------check if either or both work

Answer by timofer(155) About Me  (Show Source):
Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.
i need help finding x in the equation: 2(squareroot x) + 3 = x
~~~~~~~~~~~~~~~~~~~~~


        In the post by @mananth, the solution is incomplete and is written inaccurately;
        the answer absents, and from his post, a reader can make wrong conclusion.

        So, I came to present an accurate and complete solution.


2sqrtx +3=x
2sqrtx=x-3
square both the sides of the equation
4x=(x-3)^2
4x=x^2-6x+9
x^2-4x-6x+9=0
x^2-10x+9=0
x^2-9x-x+9=0
x(x-9)-1(x-9)=0
(x-1)(x-9)=0

The roots to the last equation are x= 1 and x= 9.

The check shows that x= 9 satisfies the original equation,
while x= 1 is an extraneous solution and should be rejected.

ANSWER. The given equation has a unique solution x = 9.

Solved completely and accurately (as it should be done).




Question 520404: square root of 12 divided by square root of 6
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
square root of 12 divided by square root of 6
*********************************************
The other person's answer, 1.414, is NOT what this author thinks, is needed!

highlight%28sqrt%2812%29%2Fsqrt%286%29%29 = sqrt%2812%29%2Fsqrt%286%29*sqrt%286%29%2Fsqrt%286%29 = 

OR

a MUCH. MUCH EASIER simplification: highlight%28sqrt%2812%29%2Fsqrt%286%29%29 = sqrt%2812%2F6%29 = sqrt%282cross%2812%29%2Fcross%286%29%29 = highlight%28sqrt%282%29%29


Question 97658: I need help completing the square for this problem.
16x^2-16x-5=0

So far I have tried this:
16x^2-16x+[1/2(-16)]=5+[1/2(-16)]
16x^2-16x+64=5+64
16(x^2-x+4)=69
16(????????????

I'm lost from this point on. I tried factoring but it's not working for me. I hope you can help me.
Thank you very much
Solve a Quadratic Equation by completing the square.
I was not sure what would be an acceptable answer for your teacher so I worked it all the way out. This problem is a Quadratic Equation. Here is your solution.

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
I need help completing the square for this problem.

16x^2-16x-5=0

So far I have tried this:

16x^2-16x+[1/2(-16)]=5+[1/2(-16)]
16x^2-16x+64=5+64
16(x^2-x+4)=69
16(????????????
 
I'm lost from this point on. I tried factoring but it's not working for me. I hope you can help me.

Thank you very much

Solve a Quadratic Equation by completing the square. 

I was not sure what would be an acceptable answer for your teacher so I worked it all the way out.
This problem is a Quadratic Equation. Here is your solution.
***********************************************************
You would be lost, because you should've FIRST divided through, by 16, in order to make the coefficient on x%5E2, 1.
16x%5E2+-+16x+-+5+=+0
This quadratic can be solved by FACTORING. Often-times, this author will solve, by FACTORING, if possible, 
before or after completing the square, and then match the solutions. You can do the same, if you wish!

           16x%5E2+-+16x+-+5+=+0
         16x%5E2%2F16+-+16x%2F16+-+5%2F16+=+0%2F16----- Dividing each side by 16
                 x%5E2+-+x+-+5%2F16+=+0 
                         x%5E2+-+x+=+5%2F16 ----- Adding 5%2F16 to both sides
x%5E2+-+x+%2B+%28%281%2F2%29%28-+1%29%29%5E2+=+5%2F16+%2B+%28%281%2F2%29%28-+1%29%29%5E2 ---- Squaring 1%2F2 of b, then adding result to both sides
          x%5E2+-+x+%2B+%28-+1%2F2%29%5E2+=+5%2F16+%2B+%28-+1%2F2%29%5E2 
                    %28x+-+1%2F2%29%5E2+=+5%2F16+%2B+1%2F4 
                   %28x+-+1%2F2%29%5E2+=+5%2F16+%2B+4%2F16 
                  %28x+-+1%2F2%29%5E2+=+9%2F16 
            sqrt%28%28x+-+1%2F2%29%5E2%29+=+0+%2B-+sqrt%289%2F16%29 ---- Taking square root on both sides
                       x+-+1%2F2+=+0+%2B-+3%2F4 
                            x+=+1%2F2+%2B-+3%2F4
                            x+=+2%2F4+%2B-+3%2F4 
                   


Question 265331: the squareroot of -12 + 2x = the squareroot of 3x+23
please provide step by step

Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.
the squareroot of -12 + 2x = the squareroot of 3x+23
please provide step by step
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution in the post by @mananth is fatally wrong.
        See my correct solution below.


Our starting equation is

    sqrt%282x-12%29 = sqrt%283x%2B23%29.


Take the square on both sides of the equation

    2x - 12  = 3x + 23

    -12 - 23 = 3x - 2x

       x    =    -35


But x = -35 makes the expression  2x-12  negative under the square root in the left side of the original equation.


So, we conclude that given equation has no real solution.

Solved correctly.




Question 969387: sir what is the root of 5+2*root 6

Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


Consider this:

%28sqrt%28x%29%2Bsqrt%28y%29%29%5E2=%28x%2By%29%2B2sqrt%28xy%29

In the form on the right, the rational part is the sum of two integers and the expression under the radical is the product of those two integers, and there is a multiplier "2" outside the radical.

In your problem, you are to find

sqrt%285%2A2sqrt%286%29%29

This is in the form of the pattern above: 5 is the sum of 2 and 3; 6 is the product of 2 and 3; and the radical has a multiplier "2". So the problem fits the pattern:

sqrt%285%2B2sqrt%286%29%29=sqrt%283%29%2Bsqrt%282%29

Here are a couple of random examples to help you see the pattern:

sqrt%2817%2B2sqrt%2870%29%29=sqrt%2810%29%2Bsqrt%287%29 [17 is 10+7; 70 is 10*7]

sqrt%289%2B2sqrt%2814%29%29=sqrt%287%29%2Bsqrt%282%29 [9 is 7+2; 14 is 7*2]

ANSWER: sqrt%285%2B2sqrt%286%29%29=sqrt%283%29%2Bsqrt%282%29


Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
sir what is the root of 5+2*root 6
**********************************root of 5 + 2*root 6
I take it, this is: sqrt%285+%2B+2sqrt%286%29%29, the square root of a SURD! If so,
                    sqrt%283+%2B+2+%2B+2sqrt%283%2A2%29%29 ----- Changing 5 to 3 + 2, and 6 (in sqrt%286%29) to 3*2
                    sqrt%283+%2B+2+%2B+2sqrt%283%29sqrt%282%29%29 ------ Applying sqrt%28m%2An%29+=+sqrt%28m%29sqrt%28n%29
  sqrt%28%28sqrt%283%29%29%5E2+%2B+%28sqrt%282%29%29%5E2+%2B+2sqrt%283%29sqrt%282%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%283%29%2C+b%2C+being%2C+sqrt%282%29%29%29, and so:
sqrt%28%28sqrt%283%29%29%5E2+%2B+%28sqrt%282%29%29%5E2+%2B+2sqrt%283%29sqrt%282%29%29 then becomes: sqrt%28%28sqrt%283%29+%2B+sqrt%282%29%29%5E2%29 
                                                                               highlight%28sqrt%283%29+%2B+sqrt%282%29%29 ----- Cancelling SQUARE and SQUARE ROOT


Question 988996: √(7-4√3) + √(7+4√3)
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


(1) In general, to find the square root of an expression of the form sqrt%28a%2Bb%2Asqrt%28c%29%29

Consider this:

%28sqrt%28x%29%2Bsqrt%28y%29%29%5E2=x%2B2sqrt%28xy%29%2By=%28x%2By%29%2B2sqrt%28xy%29 [1]

Consider the form of the expression on the right above. This required form has the sum of two integers as the rational part and the product of those two integers as the radicand, with multiplier 2 on the radical.

If we can put our given square root expression in that form, then the equivalent expression is sqrt%28x%29%2Bsqrt%28y%29

In this problem, we have the expression

sqrt%287%2B4sqrt%283%29%29

To put this in the form in [1] above, we take 2 out of the 4 and put it back inside the radical, leaving the required "2" outside the radical:

sqrt%287%2B4sqrt%283%29%29=sqrt%287%2B2sqrt%2812%29%29

Then we see that, since 4+3=7 and 4*3=12, the expression is equivalent to sqrt%284%29%2Bsqrt%283%29 or 2%2Bsqrt%283%29

So we have sqrt%287%2B4sqrt%283%29%29=2%2Bsqrt%283%29.

Similarly, we will find sqrt%287-4sqrt%283%29%29=2-sqrt%283%29.

And so the given expression is equivalent to %282%2Bsqrt%283%29%29%2B%282-sqrt%283%29%29=4

(2) And for an expression in the exact form of the given expression, there is a very different way to evaluate the expression.

Square the given expression and watch how it simplifies.

%28sqrt%287%2B4sqrt%283%29%29%2Bsqrt%287-4sqrt%283%29%29%29%5E2



14%2B2sqrt%2849-48%29=14%2B2=16

Then, since the square of the given expression is 16, the given expression is equal to 4.


Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
√(7-4√3) + √(7+4√3)
*******************
                         sqrt%287+-+4sqrt%283%29%29+%2B+sqrt%287+%2B+4sqrt%283%29%29
                 sqrt%287+-+2%282%29sqrt%283%29%29+%2B+sqrt%287+%2B+2%282%29sqrt%283%29%29
                     sqrt%287+-+2sqrt%284%29sqrt%283%29%29+%2B+sqrt%287+%2B+2sqrt%284%29sqrt%283%29%29
                 ----- Changing 7 to 4 + 3                 
sqrt%28%28sqrt%284%29%29%5E2+%2B+%28sqrt%283%29%29%5E2+-+2sqrt%284%29sqrt%283%29%29 + sqrt%28%28sqrt%284%29%29%5E2+%2B+%28sqrt%283%29%29%5E2+%2B+2sqrt%284%29sqrt%283%29%29 ------ Converting 
The above is in the form:
                          %28a+-+b%29%5E2  + %28a+%2B+b%29%5E2, with system%28matrix%282%2C3%2C+a%2C+being%2C+sqrt%284%29%2C+b%2C+being%2C+sqrt%283%29%29%29, and so:
sqrt%28%28sqrt%284%29%29%5E2+%2B+%28sqrt%283%29%29%5E2+-+2sqrt%284%29sqrt%283%29%29 + sqrt%28%28sqrt%284%29%29%5E2+%2B+%28sqrt%283%29%29%5E2+%2B+2sqrt%284%29sqrt%283%29%29 then becomes:
                   sqrt%28%28sqrt%284%29+-+sqrt%283%29%29%5E2%29 +  sqrt%28%28sqrt%284%29+%2B+sqrt%283%29%29%5E2%29
                         sqrt%284%29+-+sqrt%283%29    +    sqrt%284%29+%2B+sqrt%283%29 ----- Cancelling SQUARE and SQUARE ROOT
                               2+-+sqrt%283%29+%2B+2+%2B+sqrt%283%29+=+highlight%284%29%29


Question 967135: 2 to the n power multiplied by 2' all over 2 to -5 power equals 2 squared, solve for n
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
2 to the n power multiplied by 2' all over 2 to -5 power equals 2 squared, solve for n
*************************************2 to the n power multiplied by 2' all over 2 to -5 power equals 2 squared, solve for n
               If %282%5En+%2A+2%29%2F2%5E%28-+5%29+=+2%5E2, then:
                 %282%5En+%2A+2%5E1%29%2F2%5E%28-+5%29+=+2%5E2
              %282%5E%28n+%2B+1%29%29%2F2%5E%28-+5%29+=+2%5E2
                 2%5E%28n+%2B+1%29+=+%282%5E2%29%282%5E%28-+5%29%29 ---- Cross-multiplying
                 2%5E%28n+%2B+1%29+=+2%5E%282+%2B+-+5%29
                    n + 1 = 2 + - 5 --- EQUATING exponents, since bases are equal
                    n + 1 = - 3 
                           n = - 3  -  1 = - 4


Question 973570: Hi,
Well I've run into a problem that I can't really do.
The question is to rationalise 1/(cube root 2 - 1). The answer is cube root 2 + cube root 4 + 1. I know how to rationalise 1/cube root 2 but I'm not sure how to do this one. What I got is:
1/cube root 2-1=1/cube root 2-1 * (cube root 2^2 + 1)/(cube root 2^2 + 1)
And I've so on and so forth, but I didn't get the right answer. I got cube root 4 + 1.
Thanks

Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


Use this factoring pattern:

a%5E3-b%5E3=%28a-b%29%28a%5E2%2Bab%2Bb%5E2%29

With a=root%283%2C2%29 and b=1, this becomes

2-1=%28root%283%2C2%29-1%29%28root%283%2C4%29%2Broot%283%2C2%29%2B1%29

Your expression is

1%2Froot%283%2C2%29

To rationalize the denominator, you need to multiply numerator and denominator by

%28root%283%2C4%29%2Broot%283%2C2%29%2B1%29

Then



... which is the answer you say you are supposed to get


Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Hi,

Well I've run into a problem that I can't really do.

The question is to rationalise 1/(cube root 2 - 1). The answer is cube root 2 + cube root 4 + 1. I know
how to rationalise 1/cube root 2 but I'm not sure how to do this one. What I got is:

1/cube root 2-1=1/cube root 2-1 * (cube root 2^2 + 1)/(cube root 2^2 + 1)

And I've so on and so forth, but I didn't get the right answer. I got cube root 4 + 1.

Thanks
*************************************************
If this is 1%2F%28root+%283%2C+2%29+-+1%29, then your 1st answer, "cube root 2 + cube root 4 + 1," is WRONG, but the 2nd, "cube
root 4 + 1," is CORRECT!! See BELOW!!

1%2F%28root+%283%2C+2%29+-+1%29 = 1%2F%28-+1+%2B+root+%283%2C+2%29%29 
1%2F%28-+1+%2B+root+%283%2C+2%29%29 * %28-+1+-+root+%283%2C+4%29%29%2F%28-+1+-+root+%283%2C+4%29%29 ---- Rationalizing denominator by MULTIPLYING numerator and denominator by -+1+-+root+%283%2C+2%5E2%29 = -+1+-+root+%283%2C+4%29 
 =  = %28-+1+-+root+%283%2C+4%29%29%2F%28%28-+1%29%5E2+-+root+%283%2C+2%2A4%29%29 = %28-+1+-+root+%283%2C+4%29%29%2F%28%28-+1%29%5E2+-+root+%283%2C+8%29%29 = %28-+1+-+root+%283%2C+4%29%29%2F%281+-+2%29 = %28-+1+-+root+%283%2C+4%29%29%2F%28-+1%29 = highlight%281+%2B+root+%283%2C+4%29%29%29


Question 973174: x^1/2 + y = 7
x + y^1/2 = 11
Find the value of x and y

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
x^1/2 + y = 7
x + y^1/2 = 11
Find the value of x and y
*************************
matrix%282%2C1%2C+%22+%22%2C+x%5E%281%2F2%29+%2B+y+=+7%29____sqrt%28x%29+%2B+y+=+7 
                                   sqrt%28x%29+=+7++-++y
                              %28sqrt%28x%29%29%5E2+=+%287++-++y%29%5E2
                                     x+=+49++-++14y+%2B+y%5E2 ----- eq (i)

                              x+%2B+sqrt%28y%29+=+11
                                     x+=+11+-+sqrt%28y%29 ----- eq (ii)

We then get: 49++-++14y+%2B+y%5E2+=+11+-+sqrt%28y%29
         y%5E2+%2B+sqrt%28y%29+-+14y+%2B+49+-+11+=+0
                y%5E2+%2B+sqrt%28y%29+-+14y+%2B+38+=+0

                                 Let sqrt%28y%29+=+t
              Then: system%28%28sqrt%28y%29%29%5E2+=+t%5E2%2C+y+=+t%5E2%2C+y%5E2+=+%28t%5E2%29%5E2+=+t%5E4%29
                        y%5E2+%2B+sqrt%28y%29+-+14y+%2B+38+=+0 then becomes: 
                        t%5E4+%2B+t+-+14t%5E2+%2B+38+=+0
                        t%5E4+-+14t%5E2+%2B+t+%2B+38+=+0
Using the RATIONAL ROOT THEOREM, we find that a root of the above equation is: t = 2, which makes its
FACTOR, t - 2. When divided by t - 2, using LONG DIVISION of POLYNOMIALS, or using SYNTHETIC DIVISION,
the other factor of t%5E4++-++14t%5E2+%2B+t+%2B+38, besides t - 2, is: t%5E3+%2B+2t%5E2+-+10t+-+19.
From this, we find another REAL solution being approximately 3.13131. The other 2 are negative (< 0) and
so, MUST be REJECTED/IGNORED, since sqrt%28y%29+=+t CANNOT have a negative (< 0) value for t. 

I will continue with the REAL INTEGER value, 2.
     sqrt%28y%29+=+t+=+2 ---- Back-substituting t = 2 for sqrt%28y%29
%28sqrt%28y%29%29%5E2+=+y+=+t%5E2+=+2%5E2+=+4

       x+=+11+-+sqrt%28y%29 ----- eq (ii)
        x = 11 - 2 ----- Substituting 2 for sqrt%28y%29 in eq (ii)
        x = 9

So, the ONLY INTEGER-solution set is: (x, y) = (9, 4). I'll let you substitute the other REAL VALUE, 3.13131
for t, to determine the other SOLUTION-SET.


Question 1017156: I need help.
simplify. do not evaluate exponential numbers.
a) (3^-2/4^3)^-1 * (3/4)
b) 3^-1*2^2*3^3 / 2^2*3^2 + (3/2)^0
c) (7^4*6^3*7^-5/7^-3*6^0)^2 / (6^5*7^-2/6^-4)^3

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
I need help.  

simplify. do not evaluate exponential numbers.

a) (3^-2/4^3)^-1 * (3/4)

b) 3^-1*2^2*3^3 / 2^2*3^2 + (3/2)^0

c) (7^4*6^3*7^-5/7^-3*6^0)^2 / (6^5*7^-2/6^-4)^3
************************************************
Responding to c), the SEEMINGLY most complex of the 3.

c)  =  = %287%5E%28-+1%29%2A6%5E3%2F7%5E%28-+3%29%29%5E2%2F%286%5E5%2A7%5E-2%2F6%5E%28-+4%29%29%5E3 = %287%5E%28-+1+-+-+3%29+%2A+6%5E3%29%5E2%2F%286%5E%285+-+-+4%29+%2A+7%5E%28-+2%29%29%5E3 = %287%5E%28-+1+%2B+3%29+%2A+6%5E3%29%5E2%2F%28+6%5E%285+%2B+4%29+%2A+7%5E%28-+2%29%29%5E3
       %287%5E2+%2A+6%5E3%29%5E2%2F%286%5E9+%2A+7%5E%28-+2%29%29%5E3 = %287%5E4+%2A+6%5E6%29%2F%286%5E27+%2A+7%5E%28-+6%29%29 = 7%5E%284+-+-+6%29+%2A+6%5E%286+-+27%29 = 7%5E%284+%2B+6%29%2A6%5E%28-+21%29 = highlight%287%5E10%2A6%5E%28-+21%29%29, or highlight%287%5E10%2F6%5E21%29


Question 1051380: simplify the expression and show check of work please if you can. x and y represent nonegative real numbers. I attempted to do this expression however I have two different answers. can someone please help
sqrt(25x^32 y^8 z^6)

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
simplify the expression and show check of work please if you can.  x and y represent nonegative real numbers.
I attempted to do this expression however I have two different answers.  can someone please help  

sqrt(25x^32 y^8 z^6)
********************
Why didn't you show the "two different answers" you came up with? Wouldn't you then be able to see where you went
wrong, if you ever did?. 

sqrt%2825x%5E32y%5E8z%5E6%29
sqrt%285%5E2%28x%5E16%29%5E2%28y%5E4%29%5E2%28z%5E3%29%5E2%29 = highlight%285x%5E16y%5E4z%5E3%29


Question 991238: What is the equation to find the 21th number of the sequence: 4, 6, 10, 16, ...
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
What is the equation to find the 21th number of the sequence: 4, 6, 10, 16, ...
******************************************************************************
What is the equation to find the 21th number of the sequence: 4, 6, 10, 16, ...
The 2nd DIFFERENCES (2) are the same, so we have a QUADRATIC sequence. We then use 
the quadratic form of an equation to find the required equation. 
QUADRATIC equation form: y+=+Ax%5E2+%2B+Bx+%2B+C
We can use any 3 points. 1st point is 4, so coordinate point is (x1, y1) = (1, 4)
                                            2nd point is 6, so coordinate point is (x2, y2) = (2, 6)
                                            3rd point is 10, so coordinate point is (x3, y3) = (3, 10)

                   (1, 4)                                               (2, 6)                                                        (3, 10)
       system%28y+=+Ax%5E2+%2B+Bx+%2B+C%2C%0D%0A4+=+A%281%29%5E2+%2B+B%281%29+%2B+C%29                   system%28y+=+Ax%5E2+%2B+Bx+%2B+C%2C%0D%0A6+=+A%282%29%5E2+%2B+B%282%29+%2B+C%29                             
    4 = A + B + C ---- eq (i)            6 = 4A + 2B + C ---- eq (ii)                  10 = 9A + 3B + C ---- eq (iii)        

      4 =  A +    B + C --- eq (i)
      6 = 4A + 2B + C --- eq (ii)
    10 = 9A + 3B + C --- eq (iii) 
      2 = 3A + B ------ Subtracting eq (i) from eq (ii) ---- eq (iv)
      4 = 5A + B  ----- Subtracting eq (ii) from eq (iii) --- eq (v)
      2 = 2A ----- Subtracting eq (iv) from eq (v)
      highlight%282%2F2+=+1+=+A%29

      2 = 3(1) = B ------ Substituting 1 for A in eq (iv)
2 - 3 = B
   - 1 = B

     4 = 1 + - 1 + C ---- Substituting 1 for A, and - 1 for B, in eq (i)
     4 = C

With A being 1, B being - 1, and C being 4, the equation for this sequence is:
y+=+Ax%5E2+%2B+Bx+%2B+C = highlight%28y+=+x%5E2+-+x+%2B+4%29

So, the 21st term in this sequence, or highlight%28y%5B21%5D%29+=+21%5E2+-+21+%2B+4+=+441+-+21+%2B+4+=+highlight%28424%29%29


Question 1154852: Bobby just received an award for employee of the year and wants to frame the plaque in b) such a way as it creates a border of uniform width around the rectangular plaque. He only has enough material to create a total of 16 square inches in area for the border. If the plaque is 6 inches by 9 inches, how wide should the border be?
Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.
Bobby just received an award for employee of the year and wants to frame the plaque in b) such a way as it creates
a border of uniform width around the rectangular plaque. He only has enough material to create a total of 16 square
inches in area for the border. If the plaque is 6 inches by 9 inches, how wide should the border be?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        The solution in the post by @mananth is incomplete/unreadable.
        Strictly saying, an explanation absents in the @mananth's post.
        See below my complete solution.


Let 'x' be the uniform width of the frame.

Write an equation for the frame area

    (6+2x)*(9+2x) - 6*9 = 16


Simplify

    54 + 18x + 12x + 4x^2 - 54 = 16,

    4x^2 + 30x - 16 = 0

    2x^2 + 15x - 8 = 0


Solve using the quadratic formula:  x = 1/2 or -8.

Use the positive value.


ANSWER.  The uniform width of the frame is 1/2 of an inch.

Solved.




Question 325791: sqrt(4x+5) = 2 + sqrt(2x-1)
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!

sqrt(4x+5) = 2 + sqrt(2x-1)
===========================
I wonder why the other person chose to write a novel for this problem!
sqrt%284x+%2B+5%29+=+2+%2B+sqrt%282x+-+1%29, with x%3E=1%2F2
%28sqrt%284x+%2B+5%29%29%5E2+=+%282+%2B+sqrt%282x+-+1%29%29%5E2 --- Squaring both sides of equation
4x+%2B+5+=+4+%2B+4sqrt%282x+-+1%29+%2B+%282x+-+1%29
4x+%2B+5+=+2sqrt%282x+-+1%29+%2B+2x+%2B+3
4x+-+2x+%2B+5+-+3+=+4sqrt%282x+-+1%29
2x+%2B+2+=+4sqrt%282x+-+1%29
%282x+%2B+2%29%2F2+=+%284sqrt%282x+-+1%29%29%2F2
x+%2B+1+=+2sqrt%282x+-+1%29
%28x+%2B+1%29%5E2+=+%282sqrt%282x+-+1%29%29%5E2 ---- Squaring both sides of equation
x%5E2+%2B+2x+%2B+1+=+4%282x+-+1%29
x%5E2+%2B+2x+%2B+1+=+8x+-+4
x%5E2+%2B+2x+-+8x+%2B+1+%2B+4+=+0
x%5E2+-+6x+%2B+5+=+0
(x - 5)(x - 1) = 0
x - 5 = 0 OR x - 1 = 0
x = 5 OR x = 1
Both solutions are ACCEPTABLE, as matrix%281%2C4%2C+5+%3E=+1%2F2%2C+%22%2C%22%2C+and%2C+1+%3E=+1%2F2%29


Question 99970: Please help. I have one last problem on my test due tonight. I know I am not giving you a lot of time but the person that usually helps me has ended up in the hospital and I am stuck on this last question.
Solve sqrt2x-1=sqrt4x-1 with another -1 outside the sqrt part.
Thanks if you have the time.

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Please help.  I have one last problem on my test due tonight.  I know I am not giving you a lot of time but the person that usually helps me has ended up in the hospital and I am stuck on this last question.

Solve sqrt2x-1=sqrt4x-1 with another -1 outside the sqrt part.

Thanks if you have the time.
****************************
This does break down to (2x - 1)(2x - 5)=0, as stated by the other person. However, x=-1/2 and x=2/5 are WRONG. 
Correct answers: matrix%281%2C3%2C+highlight%28x+=+1%2F2%29%2C+or%2C+highlight%28x+=+5%2F2%29%29


Question 648428: please help me solve this equation:
sqrt ( 32 ) times(multiply) sqrt ( 6 )

i am trying to simplify this equation down to a number and a square root..
my last step i got
sqrt ( 8 ) (times) sqrt ( 4 ) (times) sqrt ( 6 )
and i dont know what to do next...
thank you in advance for your help!

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
please help me solve this equation:

sqrt ( 32 ) times(multiply) sqrt ( 6 ) 
 
 i am trying to simplify this equation down to a number and a square root..
my last step i got 

sqrt ( 8 ) (times) sqrt ( 4 ) (times) sqrt ( 6 ) 
and i dont know what to do next...
thank you in advance for your help!
===================================
There's NOTHING to solve here, as this is NOT an equation. Instead, it's a RADICAL EXPRESSION that needs to be SIMPLIFIED.
sqrt+%2832%29sqrt+%286%29 
sqrt%288%2A4%29sqrt%286%29
sqrt+%288%29sqrt+%284%29sqrt+%286%29 <=== This is CORRECT. Good job!!
sqrt%288+%2A+4+%2A+6%29 <==== This is what you do next!
sqrt%28%282%5E3%29%282%5E2%29%282%2A3%29%29
sqrt%28%282%5E3%29%282%5E2%29%282%5E1%29%283%29%29
sqrt%282%5E%283+%2B+2+%2B+1%29%2A3%29
sqrt%282%5E6+%2A+3%29+=+%282%5E3%29sqrt%283%29+=+highlight%288sqrt%283%29%29


Question 51682: I have the answer in the back of the book. Still having trouble getting there.
4^2/3 * 6^2/3 * 9^2/3
Have tried squaring 4 and then taking the cubed root but don't get good numbers.
The book says answer is 36?????

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
I have the answer in the back of the book. Still having trouble getting there.

4^2/3 * 6^2/3 * 9^2/3

Have tried squaring 4 and then taking the cubed root but don't get good numbers.

The book says answer is 36?????
===============================
 = 


Question 252005: Can you please help me with this equation: 2- sqrt 7 / sqrt 3 - sqrt 2
the answer I have is (2- sqrt 7)(sqrt 3 + sqrt 2 / 5 but i don't think it is right.

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Can you please help me with this equation: 2- sqrt 7 / sqrt 3 - sqrt 2

the answer I have is (2- sqrt 7)(sqrt 3 + sqrt 2 / 5 but i don't think it is right. 
===================================================================================
There's NOTHING to solve here. This is a SIMPLIFICATION problem!!
If %282+-+sqrt%287%29%29%2F%28sqrt%283%29+-+sqrt%282%29%29, then:
 ---- RATIONALZING the DENOMINATOR by multiplying it by its congugate, sqrt%283%29+%2B+sqrt%282%29 

%28%282+-+sqrt%287%29%29%28sqrt%283%29+%2B+sqrt%282%29%29%29%2F%283+-+2%29
highlight%28%282+-+sqrt%287%29%29%28sqrt%283%29+%2B+sqrt%282%29%29%29%29 <=== You can FOIL this, if you wish!

So, your answer: (2- sqrt 7)(sqrt 3 + sqrt 2 / 5, or +%28%282+-+sqrt%287%29%29%28sqrt%283%29+%2B+sqrt%282%29%29%29%2F5 is PARTIIALLY CORRECT. Seems like 
you ADDED 3 and 2 (5) in the DENOMINATOR, instead of SUBTRACTIING 2 from 3 (1)!! Good job, anyway!!


Question 96313: 3+sqrt%28+11+%29+ * 2+sqrt%28+11+%29+
Found 2 solutions by greenestamps, MathTherapy:
Answer by greenestamps(13327) About Me  (Show Source):
You can put this solution on YOUR website!


I'm guessing that the "3" and "2" in the statement of the problem are the levels of the roots instead of just coefficients. So I read the problem as "cube root of 11 times square root of 11". Then




Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
3+sqrt%28+11+%29+ * 2+sqrt%28+11+%29+

highlight%283sqrt%2811%29+%2A+2sqrt%2811%29%29 = 3%2A2%2Asqrt%2811%29%2Asqrt%2811%29 = 6%2811%29+=+highlight%2866%29, NOT 6sqrt%2811%29 as the other person who responded, states!!


Question 1164689: c) The equation of a curve is y = squrt of 5x + y
i. Calculate the gradient of the curve at the point where x = 1
ii. A point with coordinates (x,y) moves along the curve in such a way that
the rate of increase of x has the constant value 0.03 units per second. Find
the rate of increase of y ath the instant when x = 1.
iii. Find the area enclosed by the curve, the x - axis, the y - axis and the line
y = 1.

Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.
c) The equation of a curve is y = squrt of 5x + y
i. Calculate the gradient of the curve at the point where x = 1
ii. A point with coordinates (x,y) moves along the curve in such a way that
the rate of increase of x has the constant value 0.03 units per second. Find
the rate of increase of y ath the instant when x = 1.
iii. Find the area enclosed by the curve, the x - axis, the y - axis and the line
y = 1.
~~~~~~~~~~~~~~~~~~~~~~~~~~


Twice or trice check your post.

Check and re-check; then cross-check.

Find the error (or errors).

Fix it (fix them).

Then re-post to the forum the fixed version.




Question 485853: What is 9 divided by the cubed root of 3?
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
What is 9 divided by the cubed root of 3?

The answer the other person gave, is WRONG!! 9 divided by the cubed root of 3 is NOT 243.

9%2Froot+%283%2C+3%29 
%289%2Froot+%283%2C+3%29%29+%2A+%28root+%283%2C+3%5E2%29%2Froot+%283%2C+3%5E2%29%29 = 9%2Aroot+%283%2C+3%5E2%29%2Froot+%283%2C+3+%2A+3%5E2%29 = 9%2Aroot+%283%2C+9%29%2Froot+%283%2C+3%5E3%29 = 9%2Aroot+%283%2C+9%29%2F3 = 3cross%289%29%2Aroot+%283%2C+9%29%2Fcross%283%29 = highlight%283%2Aroot+%283%2C+9%29%29


Question 308197: What is the cubed root of 4x squared divided by the cubed root of 16?
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
What is the cubed root of 4x squared divided by the cubed root of 16?

Whatever advice the other person gave, will NOT do it, as it WON'T rationalize the denominator.

root+%283%2C+4x%5E2%29%2Froot+%283%2C+16%29
 = root+%283%2C+4x%5E2+%2A+16%5E2%29%2Froot+%283%2C+16+%2A+16%5E2%29 = root+%283%2C+%282%5E2%29%282%5E8%29x%5E2%29%2Froot+%283%2C+16%5E3%29 = root+%283%2C+%282%5E9%292x%5E2%29%2Froot+%283%2C+16%5E3%29 = 2%5E3%2Aroot+%283%2C+2x%5E2%29%2F16 = cross%288%29%2Aroot+%283%2C+2x%5E2%29%2F2cross%2816%29 = highlight%28root+%283%2C+2x%5E2%29%2F2%29 


Question 457475: How do I solve and algebra problem with a squared variable? 25=16-Tsquared

Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.
.
How do I solve highlight%28cross%28and%29%29 an algebra problem with a squared variable? 25=16-Tsquared
~~~~~~~~~~~~~~~~~~~~~~~~


        @mananth incorrectly read and interpreted the problem,
        so his answer is irrelevant and incorrect.


25 = 16 - t^2

t^2 = 16 - 25

t^2 = -9

t = sqrt%28-9%29 = +/- 3*i,  where i = sqrt%28-1%29  is the imaginary unit.    ANSWER

Solved.




Question 262842: Simplify:
(sqrt(50) - 2sqrt(5))(5sqrt(2) + sqrt(20))

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Simplify:

(sqrt(50) - 2sqrt(5))(5sqrt(2) + sqrt(20))

The many calculations by the other person are TOTALLY UNNECESSARY!! 

%28sqrt%2850%29+-+2sqrt%285%29%29%285sqrt%282%29+%2B+sqrt%2820%29%29
2sqrt%285%29 = sqrt%2820%29
5sqrt%282%29 = sqrt%2850%29

So, %28sqrt%2850%29+-+2sqrt%285%29%29%285sqrt%282%29+%2B+sqrt%2820%29%29 
    %28sqrt%2850%29+-+sqrt%2820%29%29%28sqrt%2850%29+%2B+sqrt%2820%29%29 
    sqrt%2850%29%5E2+-+sqrt%2820%29%5E2 = 50 - 20 = 30


Question 983405: Cuberoot of 32.768 with steps.
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Cuberoot of 32.768 with steps.

Cuberoot of 32.768 with steps.root+%283%2C+32.768%29

I don't know if the other person's answer helps much!! Maybe it does.....who knows!!  

1) Last digit of 32.768: 8. Because .2%5E3+=+.008, last digit of root+%283%2C+32.768%29 is .2.

2) Next, IGNORE last 3 digits, .768, and focus on the 1st 2, 32.

3) 32 falls between 3%5E3+=+27 and 4%5E3+=+64. But, because 4%5E3 is TOO LARGE, we choose 3 (matrix%281%2C2%2C+from%2C+3%5E3%29).
   This gives us 3 as the 1st digit of the root%283%2C+32.768%29.

4) Cube root of 32.768 (root%283%2C+32.768%29) = 3.2


Question 236908: I am reveiwing algebra on my own in my old age. I am stumped on a problem. I need to find the 6th root of 225. I have broken the equation down into the square and the 3rd root, but cannot find the book answer which is the cube root of 15. Can you help? I am over half way through the book and do not move on until I have solved all of the odd numbered problems and checked my answers against the book answers.
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
I am reveiwing algebra on my own in my old age.  I am stumped on a problem. I need to find the 6th root of 225.
I have broken the equation down into the square and the 3rd root, but cannot find the book answer which is the cube
root of 15.  Can you help?  I am over half way through the book and do not move on until I have solved all of the
odd numbered problems and checked my answers against the book answers.

I don't see how the other person's answer helped at all, especially when you stated the answer, in RADICAL form,
and he/she used a calculator. Couldn't anyone with a calculator and a way to punch in the info. do that? 

Anyway, highlight%28root%286%2C+225%29%29  = root%286%2C+15%5E2%29 = root%286%2F2%2C+15%29 = highlight%28root%283%2C+15%29%29
OR

highlight%28root%286%2C+225%29%29 = root%286%2C+15%5E2%29 = %2815%5E2%29%5E%281%2F6%29%29 = matrix%282%2C1%2C+%22+%22%2C+15%5E%282%281%2F6%29%29%29 = matrix%282%2C1%2C+%22+%22%2C+15%5E%281%2F3%29%29 = highlight%28root%283%2C+15%29%29


Question 100682: Good Evening Tutor,
I have two questions I was wondering if you could tell me if I got them correct?
1.) Evaluate if possible: cubed(3) sqrt of -8/27 I got - 1/6 is this correct?
2.) simplify by combining like terms. sqrt of 63 - 2 sqrt of 28 + 5 sqrt of 7.
I got 7 sqrt of 7 I think this is wrong but I am not sure where I am getting messed up I keep getting the same answer.
Any help is appreciated. Thank you.

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
Good Evening Tutor, 

I have two questions I was wondering if you could tell me if I got them correct?  

1.)  Evaluate if possible:   cubed(3) sqrt of -8/27  I got - 1/6 is this correct?  

2.)  simplify by combining like terms.  sqrt of 63 - 2 sqrt of 28 + 5 sqrt of 7.  

I got 7 sqrt of 7 I think this is wrong but I am not sure where I am getting messed up I keep getting the same answer.  

Any help is appreciated.  Thank you.  

If 1.) is root+%283%2C+%28-+8%2F27%29%29, then it's root+%283%2C+%28-+%282%29%5E3%2F%283%5E3%29%29%29 = -+2%2F3
If it's root+%283%2C+sqrt%28-+8%2F27%29%29, then it's %28%28-+8%2F27%29%5E%281%2F2%29%29%5E%281%2F3%29 = %28%28-+8%2F27%29%5E%281%2F3%29%29%5E%281%2F2%29%29 = %28-+2%2F3%29%5E%281%2F2%29 = sqrt%28-+2%2F3%29 = sqrt%28-+1+%2A+%282%2F3%29%29 = sqrt%282%2F3%29i = %28sqrt%286%29%2F3%29i
OR
Is it %28sqrt%28-+8%2F27%29%29%5E3?


Question 1117168: solve Sqrt {2+sqrt {3}}
Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
solve Sqrt {2+sqrt {3}}

solve SIMPLIFY sqrt%282+%2B+sqrt%283%29%29

sqrt%282+%2B+sqrt%283%29%29 = sqrt%281%282+%2B+sqrt%283%29%29%29 = sqrt%28%282%282+%2B+sqrt%283%29%29%29%2F2%29 ----- Multiplying RADICAND, 2+%2B+sqrt%283%29 by 1 = 2%2F2
                        = sqrt%28%284+%2B+2sqrt%283%29%29%2F2%29
                        = sqrt%28%283+%2B+1+%2B+2sqrt%283%2A1%29%29%2F2%29 ----- Replacing 4 with 3 + 1, and 3 with 3*1
                        = sqrt%28%283+%2B+1+%2B+2sqrt%283%29%2Asqrt%281%29%29%2F2%29
                        =  ---- Converting 3 to %28sqrt%283%29%29%5E2, and 1 to %28sqrt%281%29%29%5E2
The numerator, %28sqrt%283%29%29%5E2+%2B+%28sqrt%281%29%29%5E2+%2B+2sqrt%283%29%2Asqrt%281%29 is in the form: %28a+%2B+b%29%5E2, with a being sqrt%283%29, and b being sqrt%281%29

                      So,  now becomes:
                          sqrt%28%28%28sqrt%283%29+%2B+sqrt%281%29%29%5E2%29%2F2%29 
                          sqrt%28%28sqrt%283%29+%2B+sqrt%281%29%29%5E2%29%2Fsqrt%282%29%29 --- Applying sqrt%28a%2Fb%29+=+sqrt%28a%29%2Fsqrt%28b%29
                          %28sqrt%283%29+%2B+sqrt%281%29%29%2Fsqrt%282%29 ----- Cancelling square and sqrt in NUMERATOR
                          %28%28sqrt%283%29+%2B+sqrt%281%29%29%2Fsqrt%282%29%29+%2A+%28sqrt%282%29%2Fsqrt%282%29%29 ------ Rationalizing DENOMINATOR
                          sqrt%282%29%28sqrt%283%29+%2B+sqrt%281%29%29%2F%28sqrt%282%29%2Asqrt%282%29%29%29 = highlight_green%28highlight%28%28sqrt%286%29+%2B+sqrt%282%29%29%2F2%29%29


Question 717026: one number is twice the another number and their product is 2048 . what are the two number
Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.
one number is twice the another number and their product is 2048 . what are the two number
~~~~~~~~~~~~~~~~~~~~~~~~~

According to the problem, one number is x, while another number is 2x and their product is 2048

    x * (2x) = 2048,

    2x^2 = 2048

    x^2 = 2048/2 = 1024

    x = sqrt(1024) = +/- 32.


ANSWER.  The numbers are 32 and 64  OR  -32 and -64.

Solved.

----------------------------

The solution and the answers in the post by @lynnlo are incorrect.
Simply ignore his post.




Question 202363: if (125/27)^x = the cubic root of 0.6 then the value of x is _______


Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13327) About Me  (Show Source):
Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.
if (125/27)^x = the cubic root of 0.6 then the value of x is . . .
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        This problem can be solved by applying brute force,  as tutor  @Theo did it.

        But it can be solved elegantly,  if to look at the numbers and to think a bit.


Notice that the ratio  125%2F27  in the left side is nothing else as  %285%2F3%29%5E3.

In the right side,  0.6 = 6%2F10 = 3%2F5.


This suggests that the left side is  %285%2F3%29%5E%283x%29,  while the right side is  %285%2F3%29%5E%28-1%2F3%29.


So, our equation is

    %285%2F3%29%5E%283x%29 = %285%2F3%29%5E%28-1%2F3%29.


This implies

    3x = -1%2F3,  which gives  x = -1%2F9.


ANSWER.  The solution to this equation is  x = -1%2F9.

Solved, and the exact solution is found.

Notice that -1%2F9 = -0.111111111111111...


Happy learning  ( ! )

This problem is nice  !     It is  highlight%28highlight%28specially%29%29  designed so that you would get the solution in  highlight%28highlight%28PRECISELY%29%29  highlight%28highlight%28this%29%29  form
and would get the taste of mathematical discovery.




Question 243853: cube root of 27w to the 12th power over 100
cube%28+27w%5E12%2F100+%29+
Explanation:
I am a volunteer math tutor. My student disagrees with the answer I came up with, so I am writing to you for any help in justifying her answer or my answer as "more correct". Here's the problem: Show your work in solving this cube root problem.
The heart of the question is my student's answer vs. my answer. Which one is more correct?
Student:
cube( (3 x 3 x 3) w^(4 x 3)/(75 + 25) )
3w4/(5 x 5 x 5) + cube( 25 )
3w4/5 + cube ( 25 ) <<< student's answer
Tutor:
cube( (3 x 3 x 3) w^(4 x 3)/(75 + 17 + 8) )
3w4/(5 x 5 x 5) + cube( 17 ) + (2 x 2 x 2)
3w4/5 + cube ( 17 ) + 2
3w4/7 + cube (17) <<< tutor's answer
Which answer will the teacher of my student consider as "correct" ?
3w4/5 + cube ( 25 ) OR 3w4/7 + cube (17)
Thank you. Rob Miller

Answer by MathTherapy(10806) About Me  (Show Source):
You can put this solution on YOUR website!
cube root of 27w to the 12th power over 100

cube%28+27w%5E12%2F100+%29+

Explanation:

I am a volunteer math tutor.  My student disagrees with the answer I came up with, so I am writing to you for any help in
justifying her answer or my answer as "more correct".  Here's the problem:  Show your work in solving this cube root problem.

The heart of the question is my student's answer vs. my answer.  Which one is more correct?

Student:
cube( (3 x 3 x 3) w^(4 x 3)/(75 + 25) )
 3w4/(5 x 5 x 5) + cube( 25 )
 3w4/5 + cube ( 25 ) <<< student's answer

Tutor:
cube( (3 x 3 x 3) w^(4 x 3)/(75 + 17 + 8) )
 3w4/(5 x 5 x 5) + cube( 17 ) + (2 x 2 x 2)
 3w4/5 + cube ( 17 ) + 2
 3w4/7 + cube (17) <<< tutor's answer

Which answer will the teacher of my student consider as "correct" ?

 3w4/5 + cube ( 25 ) OR 3w4/7 + cube (17)

Thank you.  Rob Miller

Neither, I'm afraid to say!! One error I noticed: 5x5x5+%3C%3E+75.
Plus, the denominator cannot be split the way you two did!

root+%283%2C+%2827w%5E12%29%2F100%29 
root+%283%2C+%2810%2F10%29%28%2827w%5E12%29%2F%28100%29%29%29 ----- Multiplying RADICAND by 1: 10%2F10
root+%283%2C+10%283w%5E4%29%5E3%2F10%5E3%29 = %283w%5E4%2F10%29root+%283%2C+10%2F1%29 = highlight%28%283w%5E4%2F10%29%28root+%283%2C+10%29%29%29

OR

root+%283%2C+%2827w%5E12%29%2F100%29 = root+%283%2C+%283w%5E4%29%5E3%2F100%29 = %283w%5E4%29%2Aroot%283%2C+%281%2F100%29%29 = highlight%28%283w%5E4%29%2Aroot%283%2C+10%5E%28-+2%29%29%29%29, or highlight%28%283w%5E4%29%2Amatrix%282%2C1%2C+%22+%22%2C+%2810%5E%28-+2%2F3%29%29%29%29%29


Question 558894: use the square root property to solve the equation

(1) (x+1)^2=9

(2) x^2=121

Answer by ikleyn(53748) About Me  (Show Source):
You can put this solution on YOUR website!
.


        In his post, tutor @Theo lost some solutions to given equations.
        So,  his treatment of the problem is incomplete.
        I came to bring a correct complete solutions.


- - - - - - - - Equation (1) - - - - - - - -

(x+1)^2 = 9


Take square root of both sides.  You will get

(x+1) = +/-3.


It means that we have two cases.


Case 1.  x+1 = 3.  Then  x = 3-1 = 2.

Case 2.  x+1 = -3.  Then x = -3-1 = -4.


So, equation (1) has two solutions.  They are  x = 2  and  x = -4.


Check.  Substitute these values into equation (1) to make sure that they suit perfectly.



- - - - - - - - Equation (2) - - - - - - - -

x^2 = 121


Take square root of both sides.  You will get

x = +/-11.


It means that the equation has two solutions,  x = 11 and x = -11.


CHECK. Substitute these values into equation (2) to make sure that they suit perfectly.

Solved correctly.

------------------------------------

It might seem the loss of roots is a minor issue.

                In fact,  it is not so.

Loss of roots is a failed test of understanding square roots.
So, solving this equation correctly or incorrectly is an easy way
to check if a person does understand the subject.
In 10 seconds, everything became clear.




Question 1181912: Al-khwarizmi solved all quadratic equations by reducing them to one of six standard forms, which were then easily solvable. He described the standard forms in terms of “squares”,”roots”, and “numbers”. Here are al-khwarizmis six standard forms.
1. Squares equal to roots(example:ax^2=bx)
2. Squares equal to numbers(example:ax^2=c)
3.roots equal to numbers(example:bx=c)
4.squares and roots equal to numbers (example:ax^2+bx=c)
5. Squares and numbers equal to roots (example: ax^2+c=bx)
6.roots and numbers equal to tot squares(example:ax^2=bx+c)
Which method would you use to solve each of the six forms? Why would you use that method?
Write a quadratic equation that can be reduced to one of al-khwarizmis six forms.

Answer by CPhill(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Here's a breakdown of how to solve each of Al-Khwarizmi's six forms, along with explanations:
**1. Squares equal to roots (ax² = bx)**
* **Method:** Divide both sides by *x* (assuming *x* is not zero). This simplifies the equation to *ax = b*, which can be solved directly for *x*.
* **Why:** Dividing by *x* reduces the quadratic to a linear equation, which is much easier to solve. We must consider the case where *x* = 0 separately, as it is a solution to the original equation.
**2. Squares equal to numbers (ax² = c)**
* **Method:** Divide both sides by *a*, then take the square root of both sides. Remember to consider both the positive and negative square roots. This gives *x* = ±√(c/a).
* **Why:** Taking the square root isolates *x*. The ± is crucial because both positive and negative values, when squared, can equal *c/a*.
**3. Roots equal to numbers (bx = c)**
* **Method:** Divide both sides by *b*. This gives *x = c/b*.
* **Why:** This is already a linear equation; dividing by *b* directly solves for *x*.
**4. Squares and roots equal to numbers (ax² + bx = c)**
* **Method:** Complete the square. Divide the entire equation by *a* to get *x² + (b/a)x = c/a*. Take half of the coefficient of *x* (which is *b/2a*), square it (*b²/4a²*), and add it to both sides. This creates a perfect square on the left side: *x² + (b/a)x + b²/4a² = c/a + b²/4a²*. Rewrite the left side as *(x + b/2a)² = c/a + b²/4a²*. Then take the square root of both sides, remembering the ±, and solve for *x*.
* **Why:** Completing the square transforms the quadratic into a form where taking the square root isolates *x*.
**5. Squares and numbers equal to roots (ax² + c = bx)**
* **Method:** Rearrange the equation to get it into the standard form for completing the square (ax² - bx + c = 0). Then, follow the same "completing the square" steps as in form 4.
* **Why:** Same as form 4, completing the square makes the equation solvable by taking the square root.
**6. Roots and numbers equal to squares (ax² = bx + c)**
* **Method:** Rewrite as ax² - bx - c = 0. Then, again, complete the square as in forms 4 and 5.
* **Why:** Consistent with the previous forms, completing the square is the key to isolating x.
**Example Quadratic Equation:**
The equation 3x² + 5 = 7x can be reduced to Al-Khwarizmi's form 5 (Squares and numbers equal to roots). Subtracting 7x from both sides gives 3x² - 7x + 5 = 0.


Question 1209713: Let a and b be integer such that (2 + sqrt(5))(137) = a + b sqrt(5). Compute a^2 - 5b^2.
Answer by greenestamps(13327) About Me  (Show Source):

Question 1183944: Al-Khwarizmi solved all quadratic equations by reducing them to one of six standard forms, which were then easily solvable. He described the standard forms in terms of "squares," "roots," and "numbers." Here are al-Khwarizmi's six standard forms:
squares equal to roots (Example: ax2= bx
a x squared equals space b x)
squares equal to numbers (Example: ax2= c
a x squared equals space c)
roots equal to numbers (Example: bx=c
b x equals c)
squares and roots equal to numbers (Example: ax2+bx=c
a x squared plus b x equals c)
squares and numbers equal to roots (Example: ax2+c=bx
a x squared plus c equals b x)
roots and numbers equal tot squares (Example: ax2=bx+c
a x squared equals b x plus c)
Activity Instructions
• Which method would you use to solve each of the six forms? Why would you use that method?
• Write a quadratic equation that can be reduced to one of al-Khwarizmi's six forms.

Answer by CPhill(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Let's break down each of al-Khwarizmi's six forms and discuss appropriate solution methods, along with example equations.
**1. Squares equal to roots (ax² = bx):**
* **Method:** Divide both sides by 'x' (assuming x ≠ 0) to get ax = b, then solve for x: x = b/a. We can also factor it as x(ax-b)=0, so x=0 or x=b/a.
* **Why:** This simplifies the quadratic to a linear equation, which is easy to solve.
* **Example:** 3x² = 12x => 3x = 12 => x = 4 (or x=0).
**2. Squares equal to numbers (ax² = c):**
* **Method:** Divide both sides by 'a' to get x² = c/a, then take the square root of both sides: x = ±√(c/a).
* **Why:** This isolates x², allowing us to directly find the value(s) of x using the inverse operation (square root).
* **Example:** 5x² = 20 => x² = 4 => x = ±2.
**3. Roots equal to numbers (bx = c):**
* **Method:** Divide both sides by 'b' to get x = c/b.
* **Why:** This is already a linear equation; one simple division gives the solution.
* **Example:** 7x = 21 => x = 3.
**4. Squares and roots equal to numbers (ax² + bx = c):**
* **Method:** This is the classic quadratic equation form. Use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a. Completing the square also works.
* **Why:** The quadratic formula provides a general solution for this type of equation.
* **Example:** 2x² + 5x = 12.
**5. Squares and numbers equal to roots (ax² + c = bx):**
* **Method:** Rearrange the equation to the standard quadratic form (ax² - bx + c = 0) and then use the quadratic formula: x = (b ± √(b² - 4ac)) / 2a. Completing the square also works.
* **Why:** Similar to the previous case, the quadratic formula is a direct way to find the solution(s).
* **Example:** 3x² + 4 = 8x => 3x² - 8x + 4 = 0.
**6. Roots and numbers equal to squares (ax² = bx + c):**
* **Method:** Rearrange the equation to the standard quadratic form (ax² - bx - c = 0) and use the quadratic formula: x = (b ± √(b² + 4ac)) / 2a. Completing the square also works.
* **Why:** Again, the quadratic formula provides a general solution.
* **Example:** 2x² = 5x + 3 => 2x² - 5x - 3 = 0.


Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425