find-the-solution-4x-2-1-2x-1-2-lt-2x-9-

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) 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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