The number of roots of the equation 2∣x∣^2 − 7∣x∣ + 6=0.


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Question Number 49462 by Pk1167156@gmail.com last updated on 07/Dec/18

$The number of roots of the equation$ $2 ∣ x ∣^{2} - 7 ∣ x ∣ + 6 = 0 .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

	$∣ x ∣= x w h e n x > 0$ $= - x w h e n x < 0$ $n o w c o n s i d e r x > 0$ $2 x^{2} - 7 x + 6 = 0$ $2 x^{2} - 4 x - 3 x + 6 = 0$ $2 x (x - 2) - 3 (x - 2) = 0$ $(x - 2) (2 x - 3) = 0$ $x = 2 a n d \frac{3}{2}$ $w h e n x < 0$ $2 x^{2} + 7 x + 6 = 0$ $2 x^{2} + 4 x + 3 x + 6 = 0$ $2 x (x + 2) + 3 (x + 2) = 0$ $(x + 2) (2 x + 3) = 0$ $x = - 2 a n d \frac{- 3}{2}$ $s o x = \pm 2 a n d \pm \frac{3}{2}$
	Answered by mr W last updated on 07/Dec/18

	$l e t t =∣ x ∣⩾ 0$ $\Rightarrow 2 t^{2} - 7 t + 6 = 0$ $\Rightarrow (2 t - 3) (t - 2) = 0$ $\Rightarrow t = \frac{3}{2}, 2 (b o t h > 0 \Rightarrow o k)$ $\Rightarrow x = \pm \frac{3}{2}, \pm 2$
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