First slide

Theory of equations

Question

The real values of a for which the quadratic equation $2 x^{2} - (a^{3} + 8 a - 1) x + a^{2} - 4 a = 0$ possesses roots of opposite signs are given by

Moderate

A

$a > 5$
B

$0 < a < 4$
C

$a > 0$
D

$a > 7$

Solution

The roots of the given equation will be of opposite signs if they are real and their product is negative.
i.e. Discriminant $\geq 0$ and Product of roots $< 0$
$\Rightarrow {(a^{3} + 8 a - 1)}^{2} - 8 (a^{2} - 4 a) \geq 0$ and $\frac{a^{2} - 4 a}{2} < 0$
$\begin{array}{l} \Rightarrow a^{2} - 4 a < 0 [∵ a^{2} - 4 a < 0 \Rightarrow {(a^{3} + 8 a - 1)}^{2} - 8 (a^{2} - 4 a) \geq 0] \\ \Rightarrow 0 < a < 4 \end{array}$

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