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Question Number 170673 by solomonwells last updated on 28/May/22

Commented by cortano1 last updated on 29/May/22

⇒2(2x+(√x) )= 3(√x) +2(√(4x^2 −x)) ; x>0 ⇒4x−(√x) = 2(√(4x^2 −x)) ⇒4((√x) )^2 −(√x) = 2(√x) (√(4x−1)) ⇒(√x) (4(√x)−1−2(√(4x−1)) )=0 ⇒4(√x)−1 = 2(√(4x−1)) ⇒16x+1−8(√x) = 16x−4 ⇒8(√x) = 5 ; x=((25)/(64))

$$\:\Rightarrow\mathrm{2}\left(\mathrm{2}{x}+\sqrt{{x}}\:\right)=\:\mathrm{3}\sqrt{{x}}\:+\mathrm{2}\sqrt{\mathrm{4}{x}^{\mathrm{2}} −{x}}\:;\:{x}>\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}{x}−\sqrt{{x}}\:=\:\mathrm{2}\sqrt{\mathrm{4}{x}^{\mathrm{2}} −{x}} \\ $$$$\Rightarrow\mathrm{4}\left(\sqrt{{x}}\:\right)^{\mathrm{2}} −\sqrt{{x}}\:=\:\mathrm{2}\sqrt{{x}}\:\sqrt{\mathrm{4}{x}−\mathrm{1}} \\ $$$$\Rightarrow\sqrt{{x}}\:\left(\mathrm{4}\sqrt{{x}}−\mathrm{1}−\mathrm{2}\sqrt{\mathrm{4}{x}−\mathrm{1}}\:\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{4}\sqrt{{x}}−\mathrm{1}\:=\:\mathrm{2}\sqrt{\mathrm{4}{x}−\mathrm{1}} \\ $$$$\Rightarrow\mathrm{16}{x}+\mathrm{1}−\mathrm{8}\sqrt{{x}}\:=\:\mathrm{16}{x}−\mathrm{4} \\ $$$$\Rightarrow\mathrm{8}\sqrt{{x}}\:=\:\mathrm{5}\:;\:{x}=\frac{\mathrm{25}}{\mathrm{64}}\: \\ $$

Commented by solomonwells last updated on 29/May/22

thanks so much sir

$$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$

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