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605-655 (Medium)|   Exponents|               
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tonebeeze
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 29 Oct 2010, 23:42
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A
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72% (02:03) correct 28% (02:38) wrong based on 2699 sessions

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If \(x ≠ 0\) and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

A. \(2y\)

B. \(y\)

C. \(\frac{y}{2}\)

D. \(\frac{4y^2}{1-4y}\)

E. \(-2y\)
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A
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 26 Mar 2011, 07:34
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If \(x ≠ 0\) and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

A. \(2y\)

B. \(y\)

C. \(\frac{y}{2}\)

D. \(\frac{4y^2}{1-4y}\)

E. \(-2y\)


Square the expression to get\(x^2 = 4xy - 4y^2\);

Re-arrange: \(x^2-4xy+4y^2=0\);

\((x-2y)^2=0\);

\(x=2y\).

Answer: A.
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 26 Mar 2011, 07:51
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petrifiedbutstanding
If x≠0 and x = \sqrt{4xy-4y^2} , then, in
terms of y, x =

A. 2y
B. y
C. y/2
D. -(4y^2)/(1-2y)
E. -2y

Can someone please help?
Show more

\(x = \sqrt{4xy-4y^2}\)
Square both sides;

\(x^2 = 4xy-4y^2\)
\(x^2+4y^2-4xy=0\)
\((x-2y)^2=0\) Note: \((a-b)^2=a^2+b^2-2ab\)
\((x-2y)^2=0\)
\(x-2y=0\)
\(x=2y\)

Ans: "A"
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 30 Oct 2010, 02:03
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tonebeeze
If x does not = 0 and \(x = \sqrt{4xy-4y^2}\) , then in terms of y, x =

a. 2y
b. y
c. \(y/2\)
d. \(-4^2/1-4y\)
e. -2y
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\(x = \sqrt{4xy-4y^2}\)

Square both sides

\(x^2=4xy-4y^2\)
\(x^2+4y^2-4xy=0\)
\((x-2y)^2=0\)
\(\Rightarrow x-2y=0\)
\(x=2y\)

Answer : (A)
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 21 Nov 2013, 02:16
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If x does not equal to 0 and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

A. 2y
B. y
C. y/2
D. 4y^2/(1-4y)
E. -2y

Square the expression: \(x^2 = 4xy - 4y^2\) --> \(x^2-4xy+4y^2=(x-2y)^2=0\) --> \(x=2y\).

Answer: A.
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Similar question to practice: if-x-0-and-x-root-8xy-16y-2-then-in-terms-of-y-x-140869.html
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 25 Aug 2017, 11:05
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tonebeeze
If x does not equal to 0 and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

A. 2y
B. y
C. y/2
D. 4y^2/(1-4y)
E. -2y
Show more


We can square both sides of the given equation and obtain:

x^2 = 4xy - 4y^2

x^2 - 4xy + 4y^2 = 0

(x - 2y)(x - 2y) = 0

(x - 2y)^2 = 0

x - 2y = 0

x = 2y

Answer: A
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 14 Jul 2018, 19:11
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petrifiedbutstanding
If x≠0 and \(x = \sqrt{4xy-4y^2}\), then, in terms of y, x =

A. 2y
B. y
C. y/2
D. -(4y^2)/(1-2y)
E. -2y
Show more

We first square both sides of the equation and then simplify. We have:

x^2 = 4xy - 4y^2

4y^2 - 4xy + x^2 = 0

(2y - x)^2 = 0

2y - x = 0

2y = x

Answer: A
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 03 Jan 2019, 13:22
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fluke
petrifiedbutstanding
If x≠0 and x = \sqrt{4xy-4y^2} , then, in
terms of y, x =

A. 2y
B. y
C. y/2
D. -(4y^2)/(1-2y)
E. -2y

Can someone please help?

\(x = \sqrt{4xy-4y^2}\)
Square both sides;

\(x^2 = 4xy-4y^2\)
\(x^2+4y^2-4xy=0\)
\((x-2y)^2=0\) Note: \((a-b)^2=a^2+b^2-2ab\)
\((x-2y)^2=0\)
\(x-2y=0\)
\(x=2y\)

Ans: "A"
Show more


Gladiator59


Hi Gladi,

can you please breakdown this for me

how from here \(x^2+4y^2-4xy=0\)

we got this \((x-2y)^2=0\)


and how from this \((x-2y)^2=0\)

we got this \(x-2y=0\)

thank you :)
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 03 Jan 2019, 13:26
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dave13
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petrifiedbutstanding
If x≠0 and x = \sqrt{4xy-4y^2} , then, in
terms of y, x =

A. 2y
B. y
C. y/2
D. -(4y^2)/(1-2y)
E. -2y

Can someone please help?

\(x = \sqrt{4xy-4y^2}\)
Square both sides;

\(x^2 = 4xy-4y^2\)
\(x^2+4y^2-4xy=0\)
\((x-2y)^2=0\) Note: \((a-b)^2=a^2+b^2-2ab\)
\((x-2y)^2=0\)
\(x-2y=0\)
\(x=2y\)

Ans: "A"


Gladiator59


Hi Gladi,

can you please breakdown this for me

how from here \(x^2+4y^2-4xy=0\)

we got this \((x-2y)^2=0\)


and how from this \((x-2y)^2=0\)

we got this \(x-2y=0\)

thank you :)
Show more

What number when squared gives 0? Only 0^2 = 0. So, if \((x-2y)^2=0\), then \(x-2y=0\).
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 04 Jan 2019, 08:43
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Gladiator59
dave13,

Also, (a-b)^2= a^2-2ab+b^2 is an identity which is true for any a,b

Put a = x and b = -2y ... And you'll get the first part of your question.

Also, as Bunuel sir mentioned.. x^2 = 0 .. means X=0

Hope this answers.. let me know.

Regards,
G

Posted from my mobile device
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gladiator
thanks Gladi

but where do you see here \(x^2+4y^2-4xy=0\) that a = x and b = -2y ? :?
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 04 Jan 2019, 08:51
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This is pattern recognition. When you have two perfect squares being added ( here \(x^2\) and \((2y)^2 = 4y^2\) are definitely perfect squares) you need to immediately search for the product terms 2ab ... which is present here. \(2*(x)*(-2y) = -4xy\)

This should flash like a light-bulb and will come only from drilling the identities in.

Here is attached a list of identities which you might want to try & drill in. At least the binomial ones :-)

dave13

gladiator
thanks Gladi

but where do you see here \(x^2+4y^2-4xy=0\) that a = x and b = -2y ? :?
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Attachments

Algebraic-Identities.png
Algebraic-Identities.png [ 17.5 KiB | Viewed 29108 times ]

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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 05 May 2022, 06:16
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tonebeeze
If \(x ≠ 0\) and \(x = \sqrt{4xy - 4 y^2}\), then in terms of y, x = ?

A. \(2y\)

B. \(y\)

C. \(\frac{y}{2}\)

D. \(\frac{4y^2}{1-4y}\)

E. \(-2y\)
Show more

Given: \(x = \sqrt{4xy - 4y^2}\)

Square both sides of the equation: \(x^2 = 4xy - 4y^2\)

Strategy: Since this looks like a quadratic equation, let's set it equal to 0

Add \(4y^2\) and subtract \(4xy\) to/from both sides of the equation: \(4y^2 - 4xy + x^2 = 0\)

Strategy: This looks like one of the special products \((a-b)(a-b) = a^2 - 2ab + b^2\)

Factor the left side to get: \((2y - x)(2y - x) = 0\)

At this point, we can conclude that \(2y - x = 0\), which means \(x = 2y\)

Answer: A
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If x ≠ 0 and x = (4xy - 4y^2)^(1/2), then in terms of y, x = ? 16 Sep 2025, 10:37
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