STD 11 - 4.2 Quadratic equations and inequations — JEE STD 11 Science — Question

We have,

$p(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$

$p(\sqrt{2}+\sqrt{3}+\sqrt{6})=0$

$\therefore \quad \quad x=\sqrt{2}+\sqrt{3}+\sqrt{6}$

$\Rightarrow \quad(x-\sqrt{6})^2=(\sqrt{2}+\sqrt{3})^2$

$\Rightarrow x^2-2 \sqrt{6} x+6=2+3+2 \sqrt{6}$

$\Rightarrow \quad\left(x^2+1\right)^2=2 \sqrt{6}(x+1)^2$

$\Rightarrow \quad x^4+2 x^2+1=24\left(x^2+2 x+1\right)$

$\Rightarrow x^4-22 x^2-48 x-23=0$

$\therefore p(x)=x^4-22 x^2-48 x-23=0$

$\text { Minimum value of } x=4$