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

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Question 994263: If a function p is defined by P(x)= 2x^2-x^0/7, find the value of p(-2)
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If a function p is defined by P(x)= 2x^2-x^0/7, find the value of p(-2)
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P(-2) = 2(-2)^2 - (1/7) = 8-(1/7) = 55/7
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cheers,
Stan H.
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Question 993857: Hello!
I hope anyone can answer :)
The topic is rationalizing denominators using conjugates
4sqrt(2) + 3 / 3sqrt(2) +sqrt(3)
Thankyou! :)

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4sqrt(2) + 3 / 3sqrt(2) +sqrt(3)
===================================================
Do it like this.
---------------------

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Multiply the NUM and the DEN by the conjugate of the DEN
---
I did one of these yesterday.
email via the TY note for help or to check your work.

Question 993859: Hello!
I hope anyone can answer :)
The topic is rationalizing denominators using conjugates
2 + sqrt(2) / sqrt(2) + sqrt(5)
Thankyou! :)

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-----
Multiply the NUM and the DEN by the conjugate of the DEN
---
I did one of these yesterday.

Question 993813: I am trying to solve this problem:
a^5 bc^3
_________

I am thinking that I should multiply the on top and bottom. but I don't know. Can you help me?

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note that square root(a^4b^5c) times square root(a^4b^5c) = (a^4b^5c)
also square root(a^4b^5c) = a^2b^2square root(bc)
***********************************************************************
now we multiply numerator and denominator by square root(a^4b^5c)
(a^5 bc^3) / square root(a^4b^5c) * square root(a^4b^5c) / square root(a^4b^5c) =
( (a^5 bc^3) * a^2b^2square root(bc) ) / (a^4b^5c) =
( (a^7b^3c^3square root(bc) ) / (a^4b^5c) =
( (a^3c^2square root(bc) / b^2 )

Question 993752: I'm really having a tough time with this problem, can you help me?
here is the problem:
-+
I rewrote it using prime factorization and got this:
-+
I simplified to this:
-+
I tried to solve this and got + but the answer key shows the answer as: - Can you tell me how they got that?

Found 2 solutions by MathTherapy, Alan3354:
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I'm really having a tough time with this problem, can you help me?
here is the problem:
-+
I rewrote it using prime factorization and got this:
-+
I simplified to this:
-+
I tried to solve this and got + but the answer key shows the answer as: - Can you tell me how they got that?
Your answer is correct.
For the answer key to be correct, the problem would have to be: -+, as opposed to: -+
It may've been , and not . Check the problem!! Maybe you copied it incorrectly!



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I'm really having a tough time with this problem, can you help me?
here is the problem:
-+
I rewrote it using prime factorization and got this:
-+
I simplified to this:
-+
I tried to solve this and got + but the answer key shows the answer as: - Can you tell me how they got that?
===============================
The answer key is wrong. It doens't happen often, but it does.
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Question 993742: Hello. I am having an issue solving this problem and I look to you for help. My problem is:
-+. 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:
-+ If I continue, I get:
-+
- Can you see where I went wrong? I'm not coming up with the right answer.

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i think the answer key is wrong.

here's a quick check using my calculator.

the original equation is:

6*sqrt(12) - 8*sqrt(50) + 9*sqrt(72)

the calculator says that the value of that is equal to 40.58359956

the book answers value is equal to 57*sqrt(3) - 40*sqrt()2) which is equal to 42.15835354.

the answers are not the same so the book answer has to be wrong.

your answer is equal to 12 * sqrt(3)-40 * sqrt(2)+54 * sqrt(2) which is equal to 40.58359956.

since this is the same as you get with the original equation, then it must be right.

i did the analysis and came up with the same answer that you did.

you could simplify it further to get 12 * sqrt(3) + 14 * sqrt(2).

using your calculator to do a quick check is a good way to find out if you're right or wrong.

you just match the answers using the original equation and the revised equation.

if they match, you did good.

if they don't, you didn't.

most calculators have a sqrt key.

if they don't, then you can use the exponential form instead.

for example:

6 * sqrt(12) can also be entered as 6 * (12)^(1/2).

Question 993750: Can you help me with this?
a^5bc^3 over
How would I do this? I would think that I should multiply top and bottom by . but I don't know.

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Can you help me with this?
a^5bc^3 over
How would I do this? I would think that I should multiply top and bottom by . but I don't know.
=============================
Step 1, say what you want to do.

Question 993595: Hello!
Hope anyone can help me answer this :)
1.) A—b / a sqrt(b) –b sqrt(a)
2.) ab / a sqrt(b) — b sqrt(a)
The topic is rationalizing denominators using conjugates
Hope anyone can explain it too
Thank you! :)

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1.) A—b / a sqrt(b) –b sqrt(a)
-------

Multiply NUM and DEN by the conjugate of the DEN
-->
=
----
You can multiply the 2 terms in the NUM, but that won't accomplish much.
---------------
2.) ab / a sqrt(b) — b sqrt(a)
This is very similar to #1

Question 993351: Hello :)
Hope anyone can answer this :) thank you!
The topic is rationalizing denominators using conjugates
2 sqrt(3) + sqrt(7) / 5 sqrt(3) — 2 sqrt(7)
Hope anyone can help me :)

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The topic is rationalizing denominators using conjugates
2 sqrt(3) + sqrt(7) / 5 sqrt(3) — 2 sqrt(7)
----------------
Multiply numerator and denominator by 5sqrt(3)+2sqrt(7) to get:
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[(2sqrt(3)+sqrt(7))*(5sqrt(3)+2sqrt(7)]/[25*3 - 4*7]
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= [10*3 +4sqrt(21)+5sqrt(21)+2*7]/47
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= [44 + 9sqrt(21)]/47
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Cheers,
Stan H.
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Question 992992: 3x-5y=900 ,5x-9y=1300 then x=? and y = ?
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.
.

Multiply first equation by  5  and the second equation by  -3.  Then add. You will get

-25y + 27y = 5*900 -3*1300,

2y = 4500 - 3900 = 600.

y = 300,

x = = = 800.

Answer.  x = 800,  y = 300.

Question 992872: Can I get your help? My problem is this:

I started by subtracting the 1 from both sides which gave me Now, would I add 4x^2 and x which would give me Now my equation looks like this: From here I would square both sides
I get 9x^2+3x-1-3x+1-1=5x^3+7. If I continue, 3x-1-3x+1 would cancel out which makes the equation look like this:
9x^2-1=5x^3+7. I subtract 9x^2 from both sides and I subtract 7 from both sides so I end up with -8=-4x. Then I divide by -4 and my answer would be 2. Can you verify to see if I made a mistake?

Found 2 solutions by stanbon, Alan3354:
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3x=1+sqrt(4x^2+x+7)
-----
3x-1 = sqrt(4x^2+x+7)
---
Square both sides to get:
9x^2 -6x + 1 = 4x^2 + x + 7
----
5x^2 - 7x - 6 = 0
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5x^2 - 10x+3x - 6 = 0
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5x(x-2)+3(x-2) = 0
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(x-2)(5x+3) = 0
-----
x = 2 or x = -3/5
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Cheers,
Stan H.
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Can I get your help? My problem is this:

I started by subtracting the 1 from both sides which gave me
Now, would I add 4x^2 and x which would give me *********
No, you can't do that. You can't add x^2 terms and x terms.
****************************************************************
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Square both sides

etc

Question 992767: Hello! Hope anyone can help me :)
The area of a rectangular streamer is 20 sq.m its width is 2 sqrt(2) jow long is the streamer?

Found 3 solutions by rothauserc, stanbon, Cromlix:
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Area(A) of rectangle = length * width, then
20 = length * (2*sqrt(2))
length = 20 / (2*sqrt(2))
length = 10 / sqrt(2)
multiply the numerator and denominator by sqrt(2)
length = (10*sqrt(2)) / 2 = 5*sqrt(2)
length = 5*sqrt(2) meters

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The area of a rectangular streamer is 20 sq.m its width is 2 sqrt(2) how long is the streamer?
-----
Area/width = length
-------
length = 20 m^2/(2sqrt(2)) = 10/sqrt(2) = 10sqrt(2)/2 = 5sqrt(2)m
----------------
cheers,
Stan H.

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Hi there,
Area of rectangular streamer = 20 sq m
Width is 2 sqrt 2
Area of rectangle = width x length
20 = 2 sqrt 2 x length
Length = 20/2 sqrt 2
Length = 5 sqrt 2
..........
Proof:
2 sqrt 2 x 5 sqrt 2
sqrt 8 x sqrt 50
= sqrt 400
= 20.
Hope this helps :-)

I squared off the expression 2x-3 and used the FOIL method to get this:
4x-12x+9=30-7x I continue by subtracting 4x-12x to get -8x+9=30-7x. Here's where I think I lose it.
Somewhere I'm missing something because I keep getting x=21 but the answer key says {3}

Found 3 solutions by MathTherapy, macston, jim_thompson5910:
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I squared off the expression 2x-3 and used the FOIL method to get this:
4x-12x+9=30-7x I continue by subtracting 4x-12x to get -8x+9=30-7x. Here's where I think I lose it.
Somewhere I'm missing something because I keep getting x=21 but the answer key says {3}
First error: , not: . Major error!!



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First step: Square both sides:
(2x-3)^2
.
FOIL:Multiply first terms:

.
Continued:

The domain of the original is:
<<
so -7/4 is eliminated, leaving:
.

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Square both sides

Use the rule

Use the rule

FOIL the left side. This is where you made your mistake. You should have and not

Get everything to one side.

Combine like terms.

Next you use the quadratic formula to solve for x. I'll let you finish up.

Note: don't forget to check each possible answer back in the original equation.

Question 992017: I am trying to solve an equation in which a radical is = to a variable expression but I am having problems. I would like to ask for your help.

After I square off both sides, I get
On the left of the equal sign, the squares cancel out and if I multiply (x-2)(x-2) my problem now looks like this:
4x+13=x^2-4x+4
To continue solving, do I collect all terms to the right or to the left? This is my confusion, I hope you can help me.

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Yes, you do. You do not forget properties of numbers nor properties of equality that you have already studied and learned. They are still in effect.

Question 991601: I need help simplifying radicals. Such as problems like ^3(radical sign)54
So 54 is under the radical sign.

Found 2 solutions by ikleyn, MathLover1:

Question 991482: Hello. I am having a real \hard time with rationalizing so I hope you can help. My problem is this:
5√5-3√3
------- 4√5+2√3
Do I need to multiply by the same conjugated denominator? would it be

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you are correct
***************************************
(5sqrt(5) - 3sqrt(3)) / (4sqrt(5) + 2sqrt(3))
multiply denominator and numerator by (4sqrt(5) - 2sqrt(3))
((5sqrt(5) - 3sqrt(3)) * (4sqrt(5) - 2sqrt(3))) / (16*5 - 4*3) =
(20*5 -22sqrt(15) +6*3) / 68 =
(118 - 22sqrt(15)) / 68 =
2(59 - 11sqrt(15)) /68 =
(59 - 11sqrt(15)) / 34

Question 991173: sqrt(x+1)+sqrt(x-1)=sqrt(2x+1)
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.
+ = .

Square both sides:

+ + = .

Simplify:

+ = ,

= .

Square both sides again:

= .

= = ,

= ,

x = +/- .

Only positive root suits the original equation  (remember about the term  ).

Question 991000: sqrt(4x+1)-sqrt(x-1)=2
Found 2 solutions by MathTherapy, josgarithmetic:
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sqrt(4x+1)-sqrt(x-1)=2




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Check the minus-form carefully.

, approximation. This will work, as well as the plus-form.

Question 990871: Find the distance between the two points

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.
The points are  (-1/2, )  and  (-1/2, ).

The distance is   = .

Question 990727: What is the simplest form of the equation?
http://moodle.ncvps.org/pluginfile.php/2035767/question/questiontext/2241511/7/3693120/eq_eb0481.gif

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Question 990388: Can you help me with this? Here is the problem:
I am trying to symplify this radical equation but i'm stuck. Now the answer key shows the answer as 3x^5y^6
I get where x^5 and y^6 come from but how do they come up with 3 outside and 3x inside. I hope this email makes sense.

Found 2 solutions by josmiceli, solver91311:
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--------------------------
Note that

and

----------------------------
The answer makes sense if you
look at each step. Hope it makes sense now

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Does the fact that and help?

John

My calculator said it, I believe it, that settles it

Question 989465: This is actually a Calculus 1 problem, but I am having problems with the algebra portion. So far, I have determined that I need to multiply this problem by the conjugate of BOTH the numerator and denominator, but I've never done that. Once I get the fraction all worked out and cleaned up, I know how to handle the limit portion.
Evaluate the limit of:
(((6-x)^(1/2))-2) / (((3-x)^(1/2))-1)
as x approaches 2
Here is my work:
(((6-x)^(1/2))-2)/(((3-x)^(1/2))-1) * (((6-x)^(1/2))-2)/(((6-x)^(1/2))-2)
=> (((6-x)^2)-((4)^2))/(((6-x)^(1/2))-2)/(((3-x)^(1/2))-1)
Then do I multiply out the top, and then multiply by the conjugate of the numerator?
=> ((-x^2)-12x+36)/(((6-x)^(1/2))-2)/(((3-x)^(1/2))-1) * (((3-x)^(1/2))-1)/(((3-x)^(1/2))-1)
This is where I'm really confused. How do I multiply a polynomial by the conjugate? Would the numerator be ((-x^2)(3-(x^(1/2)))-12x((3-(x^(1/2))+36((3-x)^(1/2))-x^2+12x-36?
Now what? I appreciate any and all help. Please let me know where I went wrong, and what I need to do to work out the problem.

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Well, to answer your question straight-up, The conjugate of a binomial expression is simply the same expression with the sign in the middle changed. And the product of a binomial and its conjugate is the difference of two squares.

What it looks like you are trying to do is to rationalize both your numerator and denominator; maybe you have an irrational (no pun intended) fear of radicals. If that is the case, I hate to be the bearer of bad tidings, but you are doomed to failure. Yes, you can rationalize your numerator, at the expense of ending up with an EUPC (that's Extremely Ugly Piece of Crap) in the denominator. OR you can rationalize your denominator, at the expense of the mathematical equivalent of pornography in the numerator. But not both. And even at that, once you have either a rational numerator or denominator, that rational expression is still going to be zero in the limit, and its ugly companion will also be zero in the limit. Life lesson here: Once you have gotten yourself into a hole, stop digging.

However, the larger question is: Why are you mucking about with multiplying by conjugates? What you have is a L'Hôpital indeterminate form . Just take the derivative of the numerator and the derivative of the denominator and try to evaluate your limit again.

John

My calculator said it, I believe it, that settles it

Question 989398: How do I solve this equation:

Found 2 solutions by josgarithmetic, rothauserc:
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This solution requires squaring twice.

.
.

Either OR .
Either of them will work.

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sqrt ( 2x-3 ) - sqrt ( x-2 ) = 1
add sqrt ( x-2 ) to both sides of =
sqrt ( 2x-3 ) = 1 + sqrt ( x-2 )
square both sides of =
2x-3 = 1 +2sqrt (x-2) + x-2
x-2 = 2sqrt (x-2)
square both sides of =
x^2-4x+4 = 4x -8
x^2 -8x +12 = 0
(x-6)(x-2) = 0
x = 6 or x = 2
**********************************
check the answers in the original equation
1) x = 6
sqrt(2*6-3) - sqrt(6-2) = 1
3 -2 = 1
1 = 1
2) x = 2
sqrt(2*2-3) - sqrt(2-2) = 1
1 - 0 = 1
1 = 1

Question 989213: 89+16a square -4a square ÷2a

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Do you have a question?

Question 989123: Please tell the way to solve this : x^2-2x > |2x-1|. Thank you
Found 2 solutions by solver91311, josgarithmetic:
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To keep yourself from getting all balled up with absolute value bars, intersections and unions of solution sets, and sundry other bumps in this particular road, the best thing to do is to get a visual idea of what the answer is going to look like before you start working out the details. Hence, graph the two sides of your inequality on the same set of coordinate axes.

.

The magenta line is the graph of the quadratic function and the red line is the graph of the absolute value function.

We want to find the intervals where the quadratic function is larger than the absolute value function.

Recalling the definition of absolute value, we can create two inequalties:

And

Let's begin with the second one and add 2x to both sides

Noting that the zeros of are , we can see that must either be less than -1 or greater than 1. But if you look back at the graph, you can see that 1 is an extraneous root since nowhere in the vicinity of 1 is the quadratic function larger than the absolute value function. Hence, for THIS HALF of the problem, the appropriate interval is . Note that the interval is open on both ends because the original inequality is strictly greater than.

Looking at the other half of the problem:

I'll leave it as an exercise for you to verify that the zeros of the corresponding quadratic equation are .

Again, we have an extraneous root. is a number between 0 and 1 where the quadratic function is everywhere less than the absolute value function. However, is the point where the two functions intersect and after which the quadratic begins to be the larger valued function. Hence, the appropriate interval for this half of the problem is . Again, the "strictly greater than" relation in the original inequality demands an open interval.

Taken together the solution set is the union of the two intervals:

John

My calculator said it, I believe it, that settles it

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The only way is to use the two possible conditions for 2x-1 and solve each resulting inequality.

2x-1 NON-NEGATIVE

Parabola concave up, inequality will be false between the roots.

2x-1 NEGATIVE

Inequality will be false between the roots.

These solutions may be combined with OR.

Question 988834: square root of 1 over 57

Question 988460: What is 5 radical 200
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idk.
what does that mean?

Question 988368: I am trying to solve this problem but my answer is not the answer in the book. Can you help me figure it out? Here is the equation:
x/(x-3) + 4/(x+5)=8x/((x-3)(x+5))
I have a solution set of{-4,3} but the answer key shows the answer as {4}

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.
Multiply each side by (x-3).
Multiply each side by (x+5).

Subtract 8x from each side.

From the domain of the original equation, we know xdoes not equal 3 or -5
(that would cause division by zero),so we are left with:
You can check by putting the value in the original equation:
Negative 4 works, positive four does not,
so there is an error in your book.
.
CHECK:
For x=4:
.

.

.

False, so 4 is NOT a solution.
.
For x=-4:

True, -4 IS our solution.

Question 988361: Can you help me solve this problem?
(x^2-23)/(x-3) + 7=0

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Can you help me solve this problem?
(x^2-23)/(x-3) + 7=0
Multiply by x-3
(x^2-23) + 7(x-3) = 0

x = -11
x = 4

Question 988089: can you help me simplify this radical expression?
the problem is cube root of 16x^4y, minus 3x times the cube root of 2xy, plus 5 times the cube root of -2xy^4
so far i have gotten 2 times the cube root of 2, times the cube root of x^4y, minus 3x times the cube root of 2 times the cube root of xy, plus 5 times times the negative cube root of 2 times the cube root of xy^4

Found 2 solutions by solver91311, Boreal:
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Note the sign change in the second step because

John

My calculator said it, I believe it, that settles it

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I am going to assume that 16x^4y is 16*x^4 *y.
Cube root of 16 is 2*2 ^1/3 (cube root of [8*2]
Term 1: 2*2^1/3 {x^4/3*y*1/3}
Term 2: -3*2^1/3 *x^(1/3)*y^(1/3)
Term 3: +5 *-2^1/3*x^1/3*y^4/3
;
common terms
2^1/3. That is multiplied by (2-3-5)
x^1/3. that is multiplied by (x)
y^1/3. That is multiplied by (y)
;
in front: 2^(1/3)*x^(1/3)*y^(1/3)* {2x-3-5y}

Question 987920: Please help me solve this equation: I believe I figured out the top part of the equation... 3(^3sqrt54) = 9 (^3sqrt2) but I'm not sure how to handle the bottom part of the equation ^3sqrt18? I believe the final solution is 3(^3sqrt3) just not sure how to get rid of the sqrt/radical in the denominator. Thanks in advance for your help!
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What does "^3" mean? x^3 is , but without something preceding ^3, your expression is nonsense.

John

My calculator said it, I believe it, that settles it

Question 987774: how to solve and simplify root 6 times the 6 root of 32
I have tried to simplify inside the roots first and got a hole number of 2 but figure out how to get the radical number.
we are provided the answer of 2 6 root of 108 but I don't understand how to get it step by step.

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.
"root 6 times the 6 root of 32" . . .

Is it   ?

If so,  then it is   = = = .

Question 987110: Perform the indicated operation: (2x+3)(x²-5x+5)