solve for x the following a. ∣x∣ + 3x −4 =0 b. ∣x∣−1 = 0 c. x^2 +3∣x∣ +2 =0


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Question Number 75932 by Rio Michael last updated on 21/Dec/19

$s o l v e f o r x t h e f o l l o w i n g$ $a . ∣ x ∣ + 3 x - 4 = 0$ $b . ∣ x ∣ - 1 = 0$ $c . x^{2} + 3 ∣ x ∣ + 2 = 0$
Commented by mathmax by abdo last updated on 21/Dec/19

$c) x^{2} + 3 ∣ x ∣ + 2 = 0 \Leftrightarrow ∣ x ∣^{2} + 3 ∣ x ∣ + 2 = 0 \Rightarrow t^{2} + 3 t + 2 = 0 w h i t t =∣ x ∣$ $s o t m u s t b e ⩾ 0 Δ = 3^{2} - 4.2 = 1 \Rightarrow t_{1} = \frac{- 3 + 1}{2} = - 1$ $t_{2} = \frac{- 3 - 1}{2} = - 2 a l l v a l u e s o f t a r e < 0 \Rightarrow n o s o l u t i o n s . .$
	Answered by benjo last updated on 21/Dec/19

	Commented by Rio Michael last updated on 21/Dec/19

	$w h y s i r$
	Commented by mr W last updated on 22/Dec/19

	$x^{2} ⩾ 0$ $∣ x ∣⩾ 0$ $x^{2} + 3 ∣ x ∣ + 2 ⩾ 2$ $t h e r e f o r e x^{2} + 3 ∣ x ∣ + 2 = 0 i s i m p o s s i b l e .$ $i . e . x^{2} + 3 ∣ x ∣ + 2 = 0 h a s n o (r e a l)$ $s o l u t i o n .$
	Commented by benjo last updated on 24/Dec/19

	$how about in complex solution sir ?$
	Commented by mr W last updated on 24/Dec/19

	$c o m p l e x s o l u t i o n :$ $l e t x = b i$ $- b^{2} + 3 ∣ b ∣ + 2 = 0$ $∣ b ∣^{2} - 3 ∣ b ∣ - 2 = 0$ $∣ b ∣= \frac{3 + \sqrt{17}}{2}$ $\Rightarrow x = \pm \frac{3 + \sqrt{17}}{2} i$
	Commented by benjo last updated on 24/Dec/19

	$thanks you sir$
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