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by sam2304 » Tue Oct 02, 2012 10:31 pm
I assume the sq.root applies to the whole term

sq.rt(x^2 y^3 + 3x^2y^3)
= sq.rt(4x^2y^3)
= (2^2 * x^2 * y^3)^1/2
= 2xy*(y)^1/2, which is the given answer.
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by wayneyau1214 » Wed Oct 03, 2012 4:59 am
√x^2y^3+3x^2y^3
=√4x^2y^3

Why don't the y's add up? I mean, if you add the x's (x^2+3x^2) then should the y's add up too? (y^3+y^3=2y^3)
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by Brent@GMATPrepNow » Wed Oct 03, 2012 6:34 am
wayneyau1214 wrote:√x^2y^3+3x^2y^3
This expression could benefit from some extra brackets.
I believe we can write it as:
sqrt[(x^2)(y^3) + (3x^2)(y^3)]
Simplify: sqrt[(4x^2)(y^3)]
Add some color: sqrt[(4x^2)(y^3)]

We have a nice rule that says, sqrt(xy) = [sqrt(x)][sqrt(y)]

When we apply this, we get: [sqrt(4x^2)][sqrt(y^3)]
Simplify to get: [2x][y sqrt(y)]
Rearrange to get: 2xy[sqrt(y)]

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by GMATGuruNY » Wed Oct 03, 2012 7:47 am
wayneyau1214 wrote:√x^2y^3+3x^2y^3

I got 2xy√2y
but the answer MGMAT gave was 2xy√y
I'm not sure if this is an error they made so I just want to see what you guys got. [/img]
Since the GMAT would provide answer choices, an alternate approach is to plug in values.
If we plug x=1/2 and y=2 into √(x²y³ + 3x²y³), we get:

√( (1/2)²2³ + 3(1/2)²2³ )

= √ ( (1/4)(8) + 3(1/4)8 )

= √ (2 + 6)

= √8

= 2√2.

Now we plug x=1/2 and y=2 into the answers to see which yields our target of 2√2.

Answer choice: 2xy√y
2xy√y = 2(1/2)(2)√2 = 2√2.
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by sam2304 » Wed Oct 03, 2012 11:13 pm
wayneyau1214 wrote:√x^2y^3+3x^2y^3
=√4x^2y^3

Why don't the y's add up? I mean, if you add the x's (x^2+3x^2) then should the y's add up too? (y^3+y^3=2y^3)
x^2y^3 is a single term and 3x^2y^3 is independent term. Its same as adding xy + 2xy = 3xy, you cannot say something like 3x2y, I assume that's what your doubt is.
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by wayneyau1214 » Thu Oct 04, 2012 5:44 am
sam2304 wrote:
x^2y^3 is a single term and 3x^2y^3 is independent term. Its same as adding xy + 2xy = 3xy, you cannot say something like 3x2y, I assume that's what your doubt is.
Got it, thanks!
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