Find roots

lets \ consider \ \\ P(x)=4x^2(x^4 -\frac{3}{2}) \\ P(x)=4x^2(x+(\frac{3}{2})^{\frac{1}{4}})(x-(\frac{3}{2})^{\frac{1}{4}})(x^2+(\frac{3}{2})^{\frac{1}{2}}) \\ roots \ are \ 0,0 ,+(\frac{3}{2})^{\frac{1}{4}},-(\frac{3}{2})^{\frac{1}{4}}

now \ putting \ P'(x) =0 \ we \ can \ find \ those \ peaks \\ P'(x)=24x^5 -12 x=0 \\ x=0 \ is \ obvious \ from \ graph \\ x_0 =(\frac{1}{2})^{\frac{1}{4}} \\ P(x_0)= \sqrt{2} - 3\sqrt{2} =-2\sqrt{2} \\ now \ adding \ a constant\ means \ dragging \ the \ graph \ along \ y \ axis \\ so \ we \ see \ exactly \ two \ roots \ are \ there

roots \ are \\ (\frac{1}{2})^{\frac{1}{4}},-(\frac{1}{2})^{\frac{1}{4}}

I was looking for all roots. Your method is nice.

There is an alternative method. Hint: trig substitution

A small error btw...

Roots are (\pm\sqrt[4]{2}),(\pm\sqrt[4]{2}i)...

Ok, I had not thought of that.

But moving on from Soumik's first substitution x²=a,

we get 2a^3-3a+\sqrt 2 = 0

Now, when you see a³ and 3a etc. , there's just the possibility that a sin 3x substitution is possible.

So letting a = \sqrt 2 \sin y and dividing by √2 throughout you get

sin 3y = 1

and the roots are then got very easily

why not?

Oh - yeah...I mistook y with a.