If y = (x – 5)^2 + (x + 1)^2 – 6, then y is least when

[Gmat_mission]

$$If\ \ y=(x-5)^2+(x+1)^2-6,\ then\ y\ is\ least\ when\ \ \ x=$$ A. -2
B. -1
C. 0
D. 2
E. None of the above

[spoiler]OA=D[/spoiler].

How can I find the correct answer? Should I try option by option? Experts, I need your help.

[Jay@ManhattanReview]

$$If\ \ y=(x-5)^2+(x+1)^2-6,\ then\ y\ is\ least\ when\ \ \ x=$$ A. -2
B. -1
C. 0
D. 2
E. None of the above

[spoiler]OA=D[/spoiler].

How can I find the correct answer? Should I try option by option? Experts, I need your help.

We have to determine the least value of y = (x - 5)^2 + (x + 1)^2 - 6.

Let's expand (x - 5)^2 + (x + 1)^2, and then analyze.

y = (x - 5)^2 + (x + 1)^2 - 6

=> x^2 - 10x + 25 + x^2 + 2x + 1 - 6

=> 2x^2 - 8x + 20

2(x^2 - 4x + 10)

2(x^2 - 2*2x + 4 + 6)

2[(x^2 - 2*2x + 2^2) + 6]

2[(x - 2)^2 + 6]

Since (x - 2)^2 is a non-negative number, it's least value is 0. Thus (x - 2)^2 = 0 => x = 2

The correct answer: D

Hope this helps!

-Jay
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[BTGmoderatorRO]

$$y=\left(x-5\right)^2+\left(x+1\right)^2-6\ =\ \left(x-5\right)\left(x-5\right)+\left(x+1\right)\left(x+1\right)-6$$
$$y=x^2-5x+25+x^2+2x+1-6=2x^2-8x+20$$
$$divide\ through\ by\ 2$$
$$y=x^2-4x+10$$
$$Therefore,\ x^2-4x+10=0$$
$$By\ u\sin g\ completing\ the\ square\ method\ $$
$$x^2-4x=-10$$
$$x^2-4x+\left(-4\right)^2=-10+\left(-4\right)^2$$
$$\left(x-4\right)^2=-10+16=6$$
$$x-4=+-\sqrt{6}$$
$$x=6.4\ or\ 1.5$$
$$by\ approximation$$
$$x=6\ or\ 2$$
$$Therefore\ y\ is\ least\ when\ x=2$$
Hence, option D is correct