Question

Obtain all roots of the polynomial $x^4-3\sqrt{2}{x}^3+3x^2+3\sqrt{2x}+4,$ if two of its roots are $\sqrt{2},and,2\sqrt{2}$

Answer 1

Solution :- given

$\Rightarrow P\left(x\right)=x^4-3\sqrt{2}{x}^3+3x^2+3\sqrt{2}x-4=0$

as Two zero is given $\sqrt{2},2\sqrt{2}$

$\therefore (x-\sqrt{2})(x-2\sqrt{2})=x^2-3\sqrt{2}x+4$

Now considering a standard quadratic equation

$a{x}^2+bx+c=0$

We can say

$(x^2-3\sqrt{2}x+4)(a{x}^2+bx+c)=p\left(x\right)$

$\Rightarrow a{x}^4-3a\sqrt{2}{x}^3+4a{x}^2+b{x}^3-3b\sqrt{2}{x}^2+4bx+c{x}^2-3c\sqrt{2}x+4c=p\left(x\right)$

$\Rightarrow a{x}^4-x^3(3a\sqrt{2}-b)+x^2(4a-3b\sqrt{2}+c)+x(4b-3c\sqrt{2})+4c=p\left(x\right)$

Now comparing with P(x) we het

$\Rightarrow 4c=-4\Rightarrow 4b-3c\sqrt{2}=3\sqrt{2}\Rightarrow 3a\sqrt{2}-b=3\sqrt{2}$

$\Rightarrow c=-1\Rightarrow 4b=3\sqrt{2}-3\sqrt{2}\Rightarrow 3a\sqrt{2}=3\sqrt{2}$

$\Rightarrow b=0\Rightarrow a=1$

$\therefore$ We can say other roots are

$\Rightarrow x^2-1=0$

$\Rightarrow x^2=1$

$\Rightarrow x=\pm 1=+1,-1$ (Ans)