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Question Number 147587 by mathdanisur last updated on 22/Jul/21

Commented by ajfour last updated on 22/Jul/21

what if we replace the r.h.s. in this question by 2x ?

$${what}\:{if}\:{we}\:{replace}\:{the}\:{r}.{h}.{s}.\:{in} \\ $$$${this}\:{question}\:{by}\:\:\mathrm{2}{x}\:? \\ $$

Answered by iloveisrael last updated on 22/Jul/21

⇒x+(√(kx+2k)) +2(√(x^2 −(kx+2k)))+x−(√(kx+2k)) = 4k^2 ⇒2x−4k^2 = −2(√(x^2 −kx−2k)) ⇒x−2k^2 =−(√(x^2 −kx−2k)) ⇒x^2 −4k^2 x+4k^4 =x^2 −kx−2k ⇒(4k^2 −k)x=4k^4 +2k ⇒(4k−1)x=4k^3 +2 ⇒x = ((4k^3 +2)/(4k−1)) ; k≠(1/4)

$$\Rightarrow\mathrm{x}+\sqrt{\mathrm{kx}+\mathrm{2k}}\:+\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} −\left(\mathrm{kx}+\mathrm{2k}\right)}+\mathrm{x}−\sqrt{\mathrm{kx}+\mathrm{2k}}\:=\:\mathrm{4k}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2x}−\mathrm{4k}^{\mathrm{2}} \:=\:−\mathrm{2}\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{kx}−\mathrm{2k}} \\ $$$$\Rightarrow\mathrm{x}−\mathrm{2k}^{\mathrm{2}} \:=−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{kx}−\mathrm{2k}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} −\mathrm{4k}^{\mathrm{2}} \mathrm{x}+\mathrm{4k}^{\mathrm{4}} =\mathrm{x}^{\mathrm{2}} −\mathrm{kx}−\mathrm{2k} \\ $$$$\Rightarrow\left(\mathrm{4k}^{\mathrm{2}} −\mathrm{k}\right)\mathrm{x}=\mathrm{4k}^{\mathrm{4}} +\mathrm{2k} \\ $$$$\Rightarrow\left(\mathrm{4k}−\mathrm{1}\right)\mathrm{x}=\mathrm{4k}^{\mathrm{3}} +\mathrm{2} \\ $$$$\Rightarrow\mathrm{x}\:=\:\frac{\mathrm{4k}^{\mathrm{3}} +\mathrm{2}}{\mathrm{4k}−\mathrm{1}}\:;\:\mathrm{k}\neq\frac{\mathrm{1}}{\mathrm{4}}\: \\ $$

Commented by mathdanisur last updated on 22/Jul/21

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

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