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Question Number 101436 by bobhans last updated on 02/Jul/20

Answered by bramlex last updated on 02/Jul/20

⇔ x^2 +30 + (√(x^2 +30)) = x^4 +x^2 set (√(x^2 +30)) = z ⇔ z^2 + z = x^4 + x^2 (x^2 +z)(x^2 −z)+(x^2 −z) = 0 (x^2 −z)(x^2 +z+1)= 0 (1) x^2 = z = (√(x^2 +30)) x^4 −x^2 −30 = 0 (x^2 −6)(x^2 +5) = 0 x = ± (√6) ; x = ± i(√5) ⇐rejected (2) x^2 +1 =−(√(x^2 +30)) ⇐ rejected The Solution is x = ± (√6) ★

$$\Leftrightarrow\:\mathrm{x}^{\mathrm{2}} +\mathrm{30}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{30}}\:=\:\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{set}\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{30}}\:=\:\mathrm{z} \\ $$$$\Leftrightarrow\:\mathrm{z}^{\mathrm{2}} \:+\:\mathrm{z}\:=\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{x}^{\mathrm{2}} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{z}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{z}\right)+\left(\mathrm{x}^{\mathrm{2}} −\mathrm{z}\right)\:=\:\mathrm{0} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{z}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}\right)=\:\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{z}\:=\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{30}} \\ $$$$\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} −\mathrm{30}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{6}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{5}\right)\:=\:\mathrm{0}\: \\ $$$$\mathrm{x}\:=\:\pm\:\sqrt{\mathrm{6}}\:;\:\mathrm{x}\:=\:\pm\:{i}\sqrt{\mathrm{5}}\:\Leftarrow\mathrm{rejected} \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{2}} +\mathrm{1}\:=−\sqrt{{x}^{\mathrm{2}} +\mathrm{30}}\:\Leftarrow\:\mathrm{rejected} \\ $$$$\mathrm{The}\:\mathcal{S}\mathrm{olution}\:\mathrm{is}\:{x}\:=\:\pm\:\sqrt{\mathrm{6}}\:\bigstar \\ $$

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