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Question:

The equation x³ + (2r + 1)x² + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other roots are real, then the minimum possible non-negative integer value of r is?

Explanation:

Let p and q be the other two real roots of the given cubic equation.

Product of the three roots of the given cubic equation = - $\frac{2}{1}$ = -2 × p × q
⇒ q = 1/p

∴ The three roots are -2, p and 1/p

Sum of the three roots of the given cubic equation = - $\frac{2\mathrm{r}+1}{1}$ = -2 + p + $\frac{1}{\mathrm{p}}$
⇒ p + $\frac{1}{\mathrm{p}}$ = -2r + 1

We know than sum of a number and its reciprocal is either less than or equal to - 2 or greater than or equal to 2.

⇒ 2 ≤ p + $\frac{1}{\mathrm{p}}$ ≤ -2

⇒ 2 ≤ -2r + 1 ≤ -2

⇒ 1 ≤ -2r ≤ -3

⇒ -1/2 ≥ r ≥ 3/2

∴ Least non-negative integral value of r is 2.

Hence, 2.