CAT 2023 Slot 1 · Quant · Question 5

Both \left| x \right| and {x}^{2} depend only on the size of x, so substitute k=\left| x \right|, where k\ge 0.

Since {x}^{2}={k}^{2}, the equation becomes 2k\left({k}^{2}+1 \right)=5{k}^{2}.

Expanding, 2{k}^{3}+2k=5{k}^{2}, which rearranges to 2{k}^{3}-5{k}^{2}+2k=0.

Taking k out as a common factor, k\left( 2{k}^{2}-5k+2 \right)=0.

One root is k=0.

The quadratic factors as \left( 2k-1 \right)\left( k-2 \right)=0, giving k=\frac{1}{2} and k=2.

Translating back, \left| x \right|=0 gives x=0; \left| x \right|=\frac{1}{2} gives x=\pm \frac{1}{2}; and \left| x \right|=2 gives x=\pm 2.

That is five real solutions in all, but only integers are wanted.

The integers among them are 0, 2 and -2, so the count is 3.

Shortcut. Substituting k=\left| x \right| removes the modulus entirely and leaves a cubic that factors on sight. Each positive root of k then contributes two values of x, while k=0 contributes just one — but the fractional root is discarded before counting.