$$If\ \ x\ge0\ \ and\ \ \ x=\sqrt{8xy-16y^2}$$ 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.
If x >= 0 and x=root(8xy-16y^2), then, in terms of y, x=?
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- EconomistGMATTutor
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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.
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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AAPL wrote:$$If\ \ x\ge0\ \ and\ \ \ x=\sqrt{8xy-16y^2}$$ then, in terms of y, x=?
A. -4y
B. 4/y
C. y
D. 4y
E. 4y^2
The OA is D.
Squaring the equation, we have:
x^2 = 8xy - 16y^2
x^2 - 8xy + 16y^2 = 0
(x - 4y)(x - 4y) = 0
x - 4y = 0
x = 4y
Answer: D
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