Question

x
${x}^{4} + 1 \div {x}^{4} = \sqrt{5 } solve \: it \: plz$

Answer 1

Hey There!!
Here we are given:

$x^4+\frac{1}{x^4} = \sqrt{5} \\ \\ \\ Let\,\, x^4=t \\ \\ \\ \implies t+\frac{1}{t}=\sqrt{5} \\ \\ \\ \implies t^2+1=\sqrt{5}t \\ \\ \\ \implies t^2-\sqrt{5}t+1=0 \\ \\ \\ \implies Discriminant \, \,D = b^2-4ac = 5-4 = 1 \\ \\ \\ \implies t=\frac{-b\pm \sqrt{D}}{2a} \\ \\ \\ \implies t = \frac{\sqrt{5}\pm 1}{2} \\ \\ \\ Since\,\,t=x^4 \\ \\ \\ x^4=\frac{\sqrt{5}\pm 1}{2} \\ \\ \\ \implies \boxed{x=\left(\frac{\sqrt{5}+1}{2}\right)^{\frac{1}{4}}} \, \, \, OR \, \, \, \boxed{x=\left(\frac{\sqrt{5}-1}{2}\right)^{\frac{1}{4}}}$


Hope it helps
Purva
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