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 = - 21 = -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 = - 2r+11 = -2 + p + 1p ⇒ p + 1p = -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 + 1p ≤ -2
⇒ 2 ≤ -2r + 1 ≤ -2
⇒ 1 ≤ -2r ≤ -3
⇒ -1/2 ≥ r ≥ 3/2
∴ Least non-negative integral value of r is 2.
Hence, 2.
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