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If x >= 0 and x=root(8xy-16y^2), then, in terms of y, x=?

This topic has 2 expert replies and 0 member replies

If x >= 0 and x=root(8xy-16y^2), then, in terms of y, x=?

Post Tue Dec 05, 2017 10:46 am
$$If\ \ x\ge0\ \ and\ \ \ x=\sqrt{8xy-16y^2}$$ then, in terms of y, x=?

A. -4y
B. 4/y
C. y
D. 4y
E. 4y^2

The OA is D.

Experts, can help me with this PS question please? Thanks.

I know that I should get the x value of the expression,

$$x^2-8xy+16y^2=0$$

But, I can't determinate it, I should factorize the expression but I can't get the correct answer. I don't know where is my error.

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GMAT/MBA Expert

Post Tue Dec 05, 2017 10:55 am
Hello AAPL.

Let's take a look here.

Your first approach is correct. From $$x=\sqrt{8xy-16y^2}$$ we get $$x^2-8xy+16y^2=0.$$ Now, using the quadratic formula we get $$x_{1,2}=\frac{-\left(-8y\right)\pm\sqrt{\left(-8y\right)^2-4\cdot1\cdot16y^2}}{2}=\frac{8y\pm\sqrt{64y^2-64y^2}}{2}=\frac{8y}{2}=4y.$$ So, the correct answer is D.

I hope this explanation may help you to clarify your doubt.

I'm available if you'd like a follow up.

Regards.

_________________
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Post Tue Dec 05, 2017 6:30 pm
A little easier:

x² - 8xy + 16y² = 0

(x - 4y)² = 0

x = 4y

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