If x is negative and y is positive, what is 4x2−4xy+y2

Question

If  x  is negative and  y  is positive, what is  \sqrt{4x^2−4xy+y^2}  ?

A.  2x−y 
 B.  2x+y 
 C.  y−2x 
 D.  2x+2 
 E.  2x−2y

Bunuel · Accepted Answer

\sqrt{4x^2−4xy+y^2}=\sqrt{(2x-y)^2}=|2x - y|

Since x is negative and y is positive, then 2x - y = negative - positive = negative. Thus, |2x - y| = -(2x - y) = y - 2x.

Answer: C.

jfranciscocuencag · Answer

Use the answers:

x = -2
y = 1

\sqrt{4(-2)^2−4(-2)(1)+1^2}  =  5

A.  2x−y  =  2(-2)−(1)  = -5 ...  discard 
 B.  2x+y  =  2(-2)+1  = -3 ...  discard 
  C.  y−2x  =  1−2(-2)  = 5 ... hold 
 D.  2x+2  =  2(-2)+2  = -2 ...  discard 
 E.  2x−2y  =  2(-2)−2(1)  = -6 ...  discard

C

manass · Answer

It is the square of (2x-y) or (y-2x) . Since y is positive and x is negative it should be (y-2x).

Ps You cannot take root of a negative number.

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LeoN88 · Answer

Concept used
(a-b)^2 = a^2 + b^2 -2ab
lxl = \sqrt{x^2}

Given expression can be reduced to l2x-yl;
Now x is -ve and y is +ve= (y-2x)  C

saukrit · Answer

The  common mistake  being tested here is that we always forget that square root is always positive  i.e.  \sqrt{y^2} = |y|

dcummins · Answer

Test x=-1 and y=1 to find 3
plug-in x=-1 and y=1 into each answer choice to find the correct answer.

Fdambro294 · Answer

1st, we can Factorize the Quadratic under the Square Root:

4(x)^2 - 4xy + (y)^2 =

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

sqrt( (2x - y)^2 )

Rule: the Square Root of a Variable Squared is equivalent to the Absolute Value of the Variable (logic: both functions yield the same result)

i.e——> sqrt( x^2 ) =

Therefore, the Square Root of the Unknown Expression (2x - y)^2 SQUARED becomes:

The Absolute Value of ——

Lastly, we are Given that x = (-)Neg. and y = (+)Pos.

This means the Quantity Inside the Modulus will be a (-)Negative Input Expression

however, since the Output of a Modulus is always NON Negative—-> the Rule to Open the Modulus is the following:

When X < 0 ———   = -(X)

Thus:

= -(2x - y) = -2x + y =

y - 2x

Answer -C-

Another way to think about it:

If X = (-)Neg. and Y = (+)

2x - y = (-)Negative Result

However, the Output of an Absolute Value Modulus can NEVER be (-)Negative. And the Answer Choices are giving us the result of an Absolute Value Expression.

Thus, 2x - y could never be the correct answer.

Posted from my mobile device

hsingh3031 · Answer

I think this question can be solved by inputting the values.

Let x = -1 and y =1

So Value is (4* 1 + 4 +1 )^2 = 9^2 = 3

So Put the values in options to get 3

Option C gives 3

OA- C

Basshead · Answer

\sqrt{4x^2−4xy+y^2}

=  (2x-y)^2 = |2x-y|

Since x is negative and y is positive =  y - 2x

Answer is C.

GmatPoint · Answer

The equation under the square root presented is:  \sqrt{\ 4x^2+y^2-4xy}

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

\left(2x-y\right)^2\ or\ \left(y-2x\right)^2

=  \left|2x-y\right|

Since x is negative and y is positive.

\left|2x-y\right|\ =-\left(2x-y\right)\ =\ y-2x

arbazfatmi1994 · Answer

The key to this question is to know that you can't take the root of a negative number.

krishna_sunder_ · Answer

check the below solution

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If x is negative and y is positive, what is 4x2−4xy+y2

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If x is negative and y is positive, what is 4x2−4xy+y2

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