If the roots of the equation $3 x^{2} + 2 (k^{2} + 1) x + (k^{2} - 3 k + 2) = 0$ be of opposite signs, then prove that 1 < k < 2

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Solution

The roots will be the opposite signs if their product is - ive and also the roots are real .
$∴ Δ \geq 0$
$p = - i v e$
$4 (k^{2} + 1)^{2} - 12 (k^{2} - 3 k 2) = + i v e$
and $\frac{k^{2} - 3 k 2}{3} = - i v e < 0$
If (2) hold then (1) automatically hold . Hence we must have $(k - 1) (k - 2) = - i v e$
$∴ 1 < k < 2$

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