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

find the solution ((4x^2 )/((1−(√(2x+1)))^2 )) < 2x+9

$$\mathrm{find}\:\mathrm{the}\:\mathrm{solution}\: \\ $$ $$\frac{\mathrm{4x}^{\mathrm{2}} }{\left(\mathrm{1}−\sqrt{\mathrm{2x}+\mathrm{1}}\right)^{\mathrm{2}} }\:<\:\mathrm{2x}+\mathrm{9} \\ $$

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)

$$\left(\mathrm{1}\right)\:\mathrm{2x}\:+\mathrm{9}\:>\:\mathrm{0}\:\Rightarrow\:\mathrm{x}\:>\:−\frac{\mathrm{9}}{\mathrm{2}} \\ $$ $$\left(\mathrm{2}\right)\:\mathrm{x}\:\geqslant\:−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ $$\left(\mathrm{3}\right)\:\mathrm{4x}^{\mathrm{2}} \:<\:\left(\mathrm{2x}+\mathrm{9}\right)\left(\mathrm{1}−\sqrt{\mathrm{2x}+\mathrm{1}}\right)^{\mathrm{2}} \\ $$ $$\mathrm{let}\:\sqrt{\mathrm{2x}+\mathrm{1}}\:=\:\mathrm{t}\:\Rightarrow\mathrm{x}\:=\:\frac{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}} \\ $$ $$\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} <\:\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{t}^{\mathrm{2}} +\mathrm{8}\right) \\ $$ $$\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \:\left\{\:\left(\mathrm{t}+\mathrm{1}\right)^{\mathrm{2}} \:−\left(\mathrm{t}^{\mathrm{2}} +\mathrm{8}\right)\:\right\}\:<\mathrm{0} \\ $$ $$\left(\mathrm{t}−\mathrm{1}\right)^{\mathrm{2}} \:\left(\mathrm{2t}−\mathrm{7}\right)\:<\mathrm{0} \\ $$ $$\Rightarrow\:\mathrm{t}\:<\:\mathrm{1}\:\cup\:\mathrm{1}\:<\:\mathrm{t}\:<\:\frac{\mathrm{7}}{\mathrm{2}} \\ $$ $$\Rightarrow\:\sqrt{\mathrm{2x}+\mathrm{1}}\:<\:\mathrm{1}\:\cup\:\mathrm{1}\:<\:\sqrt{\mathrm{2x}+\mathrm{1}}\:<\:\frac{\mathrm{7}}{\mathrm{2}} \\ $$ $$\Rightarrow\:\mathrm{x}\:<\:\mathrm{0}\:\cup\:\mathrm{0}\:<\:\mathrm{x}\:<\:\frac{\mathrm{45}}{\mathrm{8}} \\ $$ $$\mathrm{the}\:\mathrm{solution}\:\mathrm{is}\:\left(\mathrm{1}\right)\wedge\left(\mathrm{2}\right)\wedge\left(\mathrm{3}\right) \\ $$ $$−\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{x}<\mathrm{0}\:\cup\:\mathrm{0}\:<\:\mathrm{x}\:<\:\frac{\mathrm{45}}{\mathrm{8}} \\ $$

Commented byjagoll last updated on 03/Mar/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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