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

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

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Updated on: 18 Oct 2012, 05:24
3
4
00:00

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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Re: x= sqrt(8xy-16y^2), in terms of y , x = ?  [#permalink]

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17 Oct 2012, 13:28
3
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?

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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11 Dec 2012, 03:39
1
$$x=\sqrt{8xy - 16y^2}$$
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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11 Feb 2013, 18:23
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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01 Dec 2014, 01:06
$$x=\sqrt{8xy - 16y^2}$$

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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04 Apr 2018, 17:25
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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04 Sep 2019, 02:44
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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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