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Explanation:

Case 1: x ≥ 0 ⇒ |x| = x
∴ 2 × x × (x² + 1) = 5x²
⇒ 2x(x² + 1) = 5x² 
⇒ 2x(x² + 1) - 5x² = 0
⇒ x[2(x² + 1) - 5x] = 0
⇒ x(2x² – 5x + 2) = 0 
⇒ x(2x² – 4x - x + 2) = 0 
⇒ x[2x(x – 2) - (x - 2)] = 0 
⇒ x(2x - 1)(x - 2) = 0
⇒ x = 0 or ½ or 2.
We need only integral solutions hence acceptable answers are 0 and 2.

Case 2: x < 0 ⇒ |x| = -x
∴ 2 × -x × (x² + 1) = 5x²
⇒ -2x(x² + 1) = 5x² 
⇒ 2x(x² + 1) + 5x² = 0
⇒ x[2(x² + 1) + 5x] = 0
⇒ x(2x² + 5x + 2) = 0
⇒ x(2x² + 4x + x + 2) = 0
⇒ x[(2x(x + 2) + (x + 2)] = 0
⇒ x(2x + 1)(x + 2) = 0
⇒ x = 0 or -1/2 or -2
We need only integral solutions hence acceptable answers are 0 and -2.

∴ Acceptable integral solutions are -2, 0 and 2, i.e., 3 integral solutions.

Hence, 3.