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Manager  Joined: 22 Dec 2011
Posts: 211
If x>=0 and x=root(8xy - 16y^2) then, in terms of y, x=  [#permalink]

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Difficulty:   25% (medium)

Question Stats: 72% (01:41) correct 28% (02:24) wrong based on 153 sessions

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If $$x\geq{0}$$ and $$x=\sqrt{8xy - 16y^2}$$ then, in terms of y, x=

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

Originally posted by Jp27 on 17 Oct 2012, 13:13.
Last edited by Bunuel on 18 Oct 2012, 05:24, edited 2 times in total.
Renamed the topic and edited the question.
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Posts: 58452
Re: x= sqrt(8xy-16y^2), in terms of y , x = ?  [#permalink]

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Jp27 wrote:
x= sqrt(8xy-16y^2), in terms of y , x = ?

I'm sorry I dont have the answer choices with me. Could you please let me know how to solve this one?

The question should read:

If $$x\geq{0}$$ and $$x=\sqrt{8xy - 16y^2}$$ then, in terms of y, x=

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

$$x=\sqrt{8xy - 16y^2}$$, square the expression: $$x^2=8xy-16y^2$$ --> $$x^2-8xy+16y^2=0$$ --> $$(x-4y)^2=0$$ --> $$x=4y$$.

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Re: If x>=0 and x=root(8xy - 16y^2) then, in terms of y, x=  [#permalink]

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$$x=\sqrt{8xy - 16y^2}$$
Square both sides
$$x^2 = 8xy - 16y2$$
$$x^2 - 8xy + 16y^2 = 0$$

Factor the expression: $$(x - 4y)(x - 4y) = 0-->x=4y$$

Techniques in factoring: http://burnoutorbreathe.blogspot.com/2012/12/algebra-factoring-binomial-expressions.html
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4112019 wrote:
If x = 0 and x =$$\sqrt{(8xy-16y^2)}$$ , then, in terms of y, x =
A. – 4y
B.4y
C. y
D. 4y
E. 4y^2

This type of algebra highlights how important it is to recognize the difference of squares in algebra. Using this concept, this question is quite quick and simple. Without the difference of squares, you may end up with some head scratching algebra to work out. If you want to use the difference of squares, you should recognize that you need an X squared. This is your first hint to square the equation and get rid of that pesky square root sign. Moving all the variables to the same side first, you get:

$$x - \sqrt{(8xy-16y^2)} = 0$$

Changing sides to have 0 on the right as usual:

$$0 = x-\sqrt{(8xy-16y^2)}$$

Square both sides:

$$0^2 = (x-\sqrt{(8xy-16y^2)})^2$$

Since 0^2 = 0, we apply the square on the right hand side, cancelling the square root, and then distribute the minus sign to both terms:

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

Which is the difference of squares $$0 = (x-4y) (x-4y)$$

Which can be rewritten $$0 = (x-4y)^2$$

Now take the square root of both sides

$$0 = x-4y$$

So $$x = 4y$$

Answer choice B (or D since they appear to be the same in the OP question. Also of note: The first part should be x≠0)
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Re: If x>=0 and x=root(8xy - 16y^2) then, in terms of y, x=  [#permalink]

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$$x=\sqrt{8xy - 16y^2}$$

Square both sides

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

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

LHS is a perfect square of (x-4y)

$$(x-4y)^2 = 0$$

x = 4y

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Re: If x>=0 and x=root(8xy - 16y^2) then, in terms of y, x=  [#permalink]

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Jp27 wrote:
If $$x\geq{0}$$ and $$x=\sqrt{8xy - 16y^2}$$ then, in terms of y, x=

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

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

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Re: If x>=0 and x=root(8xy - 16y^2) then, in terms of y, x=  [#permalink]

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_________________ Re: If x>=0 and x=root(8xy - 16y^2) then, in terms of y, x=   [#permalink] 04 Sep 2019, 02:44
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