Question 57

What should come in place of the question mark (?) in the following equation?
$$(?)^{2} +? + 12^{3} = 13^{2} + 22^{2} + 2715$$

Solution

Let's assume 'y' in place of the question mark.

$$(y)^{2} + y + 12^{3} = 13^{2} + 22^{2} + 2715$$

$$y^2+y+1728=169+484+2715$$

$$y^2+y+1728=3368$$

$$y^2+y+1728-3368=0$$

$$y^2+y-1640=0$$

$$y^2+(41-40)y-1640=0$$

$$y^2+41y-40y-1640=0$$

$$y(y+41)-40(y+41)=0$$

(y+41) (y-40) = 0

y = -41, 40

As per the given options, '40' will come in place of the question mark.

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