The easiest IIT JEE Problem.

We first note, by completing the square, that

$y=x^{2}-8x+12 = (x-4)^{2}-4$

which reveals that the symmetry line of the parabola is $x=4$ . Furthermore, $y=0$ at $x=2$ and $x=6$ . A simple graph now shows that the square $ABCD$ is also symmetrical about $x=4$ and that point $A$ must lie somewhere in the interval $x \in <2,4>$ and $B$ in $x \in <4,6>$ .

The sides of a square have equal length, so we may conclude that the distance from the symmetry line to point $B$ will be half the distance of $BC$ :

$2 \cdot | \frac{1}{2} AB| = |BC| = 2 \cdot (x-4) = |y|$

Since the graph is under the x-axis, $|y|= -y$ , the equation for finding point $B$ becomes

$2(x-4)=-x^{2}+8x-12 \iff x^{2}-6x+4=0$

which has the solutions are $x=3 \pm \sqrt{5}$ .

The point $B$ must lie between the values $4$ and $6$ , so the only option is $x=3 + \sqrt{5}$ .

The area of $ABCD$ may now be calculated,

$( 2 \cdot ( x-4)) ^{2} = (2 \cdot (3 + \sqrt{5} -4) )^{2} = \boxed {24-8 \sqrt{5}}$

First of all given figure is completely wrong

Roots of the equation is 2 and 6

When the square is drawn it touches 2 points of the parabola below x-axis

The distance of the point from X-axis be "a", so substitute y=-a in the equation of Parabola

The roots of the equation differ by "a" so using (p-q)= $\sqrt{(p+q)^{2}-4pq}$ a quadratic in "a" is obtained

a=-2+2 $\sqrt{5}$

Area= $a^{2}$ = $\boxed {24-8\sqrt{5}}$

I think the figure is WRONG. Because the roots of the equation are 6,2.

Since the square has equal sides, this infers that the y value of the point at which the curve meets the square equals the x value of the point at which the x-axis meets the square. Hence $x = y$ , or $x = x^{2} - 8x + 12$ , implying that $0 = x^2 - 9x + 12.$ Applying the quadratic formula gives the side length of square, from which the area can be found.