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Question Number 121538 by Khalmohmmad last updated on 09/Nov/20

	Answered by benjo_mathlover last updated on 09/Nov/20
	$(i) ∣2x^2 +3x−2∣ = ∣(2x−1)(x+2)∣ → { ((2x^2 +3x−2 ; if x≤−2 ∪ x≥(1/2))),((−2x^2 −3x+2 ; if −2≤x≤(1/2))) :} (ii) ∣x−2∣ → { ((x−2 ; if x>2)),((−x+2 ; if x<2)) :} case(1) for x>2 ⇒(((2x−1)(x+2)(1−x)(1+x))/(x(x+3)(x−2))) ≤ 0 ⇒ we get solution x > 2 case(2) for x≤−2 ∪ (1/2)≤x<2 ⇒(((2x−1)(x+2)(1−x)(1+x))/(x(x+3)(2−x))) ≤ 0 we get solution x<−3 ∪ 1≤x<2 case(3)for −2≤x≤(1/2) ⇒(((1−2x)(x+2)(1−x)(1+x))/(x(x+3)(2−x))) ≤ 0 we get solution −1≤x<0 finally the solution is x>2 ∪ x<−3 ∪1≤x<2 ∪ −1≤x<0 or x<−3 ∪ −1≤x<0 ∪ x≥1 ; x≠2 (−∞,−3) ∪ [−1,0 ) ∪ [1, 2) ∪ (2,∞ )$
	$(i) ∣ {2 x}^{2} + 3 x - 2 ∣ = ∣ (2 x - 1) (x + 2) ∣$ $\to {\begin{cases} {2 x}^{2} + 3 x - 2; if x ⩽ - 2 \cup x ⩾ \frac{1}{2} \\ - {2 x}^{2} - 3 x + 2; if - 2 ⩽ x ⩽ \frac{1}{2} \end{cases}$ $(ii) ∣ x - 2 ∣ \to {\begin{cases} x - 2; if x > 2 \\ - x + 2; if x < 2 \end{cases}$ $case (1) for x > 2$ $\Rightarrow \frac{(2 x - 1) (x + 2) (1 - x) (1 + x)}{x (x + 3) (x - 2)} ⩽ 0$ $\Rightarrow we get solution x > 2$ $case (2) for x ⩽ - 2 \cup \frac{1}{2} ⩽ x < 2$ $\Rightarrow \frac{(2 x - 1) (x + 2) (1 - x) (1 + x)}{x (x + 3) (2 - x)} ⩽ 0$ $we get solution x < - 3 \cup 1 ⩽ x < 2$ $case (3) for - 2 ⩽ x ⩽ \frac{1}{2}$ $\Rightarrow \frac{(1 - 2 x) (x + 2) (1 - x) (1 + x)}{x (x + 3) (2 - x)} ⩽ 0$ $we get solution - 1 ⩽ x < 0$ $finally the solution is$ $x > 2 \cup x < - 3 \cup 1 ⩽ x < 2 \cup - 1 ⩽ x < 0$ $or x < - 3 \cup - 1 ⩽ x < 0 \cup x ⩾ 1; x \neq 2$ $(- \infty, - 3) \cup [- 1, 0) \cup [1, 2) \cup (2, \infty)$
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