Given that the solution set of the quadratic inequality ax^2 +bx+c >0 is (2,3). Then the solution set of the inequality cx^2 +bx+a <0 will be


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Question Number 162366 by cortano last updated on 29/Dec/21

$G i v e n t h a t t h e s o l u t i o n s e t o f t h e$ $q u a d r a t i c i n e q u a l i t y {a x}^{2} + b x + c > 0$ $i s (2, 3) . T h e n t h e s o l u t i o n s e t$ $o f t h e i n e q u a l i t y {c x}^{2} + b x + a < 0$ $w i l l b e$
	Answered by mr W last updated on 29/Dec/21

	$a < 0$ $x^{2} + \frac{b}{a} x + \frac{c}{a} = (x - 2) (x - 3)$ $\frac{b}{a} = - (2 + 3) = - 5 \Rightarrow b = - 5 a$ $\frac{c}{a} = 2 \times 3 = 6 \Rightarrow c = 6 a$ ${c x}^{2} + b x + a = a (6 x^{2} - 5 x + 1) < 0$ $s i n c e a < 0,$ $\Rightarrow 6 x^{2} - 5 x + 1 > 0$ $(2 x - 1) (3 x - 1) > 0$ $\Rightarrow x < \frac{1}{3} o r x > \frac{1}{2}$ $i . e . x \in (- \infty, \frac{1}{3}) \land (\frac{1}{2}, + \infty)$
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