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Question Number 171815 by Mikenice last updated on 21/Jun/22

solve for x: (2−x^2 )^(x^2 −3(√(2 )) x+4) =1

$${solve}\:{for}\:{x}: \\ $$$$\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{{x}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}\:}\:{x}+\mathrm{4}} =\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 21/Jun/22

(2−x^2 )^(x^2 −3(√(2 )) x+4) =1 case1: 2−x^2 =1⇒x=±1 case2:x^2 −3(√2) x+4=0 x=((3(√2) ±(√((−3(√2) )^2 −4(1)(4))))/(2(1))) x=((3(√2) ±(√(18−16)))/2) x=((3(√2) ±(√2))/2)=2(√2) , (√2)

$$\left(\mathrm{2}−{x}^{\mathrm{2}} \right)^{{x}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}\:}\:{x}+\mathrm{4}} =\mathrm{1} \\ $$$${case}\mathrm{1}:\:\mathrm{2}−{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}=\pm\mathrm{1} \\ $$$${case}\mathrm{2}:{x}^{\mathrm{2}} −\mathrm{3}\sqrt{\mathrm{2}}\:{x}+\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\pm\sqrt{\left(−\mathrm{3}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{4}\right)}}{\mathrm{2}\left(\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{18}−\mathrm{16}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{3}\sqrt{\mathrm{2}}\:\pm\sqrt{\mathrm{2}}}{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}}\:,\:\sqrt{\mathrm{2}} \\ $$

Commented by Mikenice last updated on 21/Jun/22

please continue your solution

$${please}\:{continue}\:{your}\:{solution} \\ $$

Commented by Mikenice last updated on 21/Jun/22

please continue your solution

$${please}\:{continue}\:{your}\:{solution} \\ $$

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