If x and y are real numbers, then the
minimum value of x

^{2}+4xy+6y^{2}-4y+4 is:
(A)-4 (B) 0 (C) 2 (D) 4
(E) None of these

Answer follows here:
Solution:

“The funda is minimum value of square
of a number is zero as it should not be negative”

So we will try to convert the given
expression into sum of squares.

x

^{2}+4xy+6y^{2}-4y+4 = x^{2}+4xy+4y^{2}+2y^{2}-4y+4
= (x

^{2}+4xy+(2y)^{2})+ ½ (4y^{2}-8y+8)
= (x+2y)

^{2}+ ½ {(2y)^{2}-2*2y*2+2^{2}+4}
= (x+2y)

^{2}+ ½ {(2y-2)^{2}+4}
= (x+2y)

^{2}+ ½ (2y-2)^{2 }+ 2
Minimum value of (x+2y)

^{2}is zero
Minimum value of (2y-2)

^{2}is zero
The remaining part is ‘2’ which is
the minimum value of the whole expression

Answer (C)
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