find the solution ((4x^2 )/((1−(√(2x+1)))^2 ))


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Question Number 83512 by jagoll last updated on 03/Mar/20

$find the solution$ $\frac{{4 x}^{2}}{{(1 - \sqrt{2 x + 1})}^{2}} < 2 x + 9$
	Answered by john santu last updated on 03/Mar/20
	$(1) 2x +9 > 0 ⇒ x > −(9/2) (2) x ≥ −(1/2) (3) 4x^2 < (2x+9)(1−(√(2x+1)))^2 let (√(2x+1)) = t ⇒x = ((t^2 −1)/2) (t^2 −1)^2 < (t−1)^2 (t^2 +8) (t−1)^2 { (t+1)^2 −(t^2 +8) } <0 (t−1)^2 (2t−7) <0 ⇒ t < 1 ∪ 1 < t < (7/2) ⇒ (√(2x+1)) < 1 ∪ 1 < (√(2x+1)) < (7/2) ⇒ x < 0 ∪ 0 < x < ((45)/8) the solution is (1)∧(2)∧(3) −(1/2)≤x<0 ∪ 0 < x < ((45)/8)$
	$(1) 2 x + 9 > 0 \Rightarrow x > - \frac{9}{2}$ $(2) x ⩾ - \frac{1}{2}$ $(3) {4 x}^{2} < (2 x + 9) {(1 - \sqrt{2 x + 1})}^{2}$ $let \sqrt{2 x + 1} = t \Rightarrow x = \frac{t^{2} - 1}{2}$ ${(t^{2} - 1)}^{2} < {(t - 1)}^{2} (t^{2} + 8)$ ${(t - 1)}^{2} {{(t + 1)}^{2} - (t^{2} + 8)} < 0$ ${(t - 1)}^{2} (2 t - 7) < 0$ $\Rightarrow t < 1 \cup 1 < t < \frac{7}{2}$ $\Rightarrow \sqrt{2 x + 1} < 1 \cup 1 < \sqrt{2 x + 1} < \frac{7}{2}$ $\Rightarrow x < 0 \cup 0 < x < \frac{45}{8}$ $the solution is (1) \land (2) \land (3)$ $- \frac{1}{2} ⩽ x < 0 \cup 0 < x < \frac{45}{8}$
	Commented by jagoll last updated on 03/Mar/20

	$thank you sir$
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