Substituting p=x^2 for Solving Polynomial Equation by Hand

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is there a way to solve this by hand
(1/2)x^4-x^2-1=0
 
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Yes substitute p=x^2.
 


(1/2)x^4-x^2-1=(1/2)p^2-p-1=0

quadratic in p

[1+||- (1+2)^(1/2) ] = 1+||-3^(1/2) = p = x^2

x=+||-[1+||-3^(1/2) ]^(1/2) =+||-[1+3^(1/2) ]^(1/2)
 


That's correct.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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