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4-2.Quadratic Equations and Inequations
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Let $p(x)=a_0+a_1 x+\ldots+a_n x^n$ be a non-zero polynomial with integer coefficients. If $p(\sqrt{2}+\sqrt{3}+\sqrt{6})=0$, then the smallest possible value of $n$ is
A
$8$
B
$6$
C
$4$
D
$2$
(KVPY-2009)
Solution
(c)
We have,
$p(x)=a_0+a_1 x+a_2 x^2+\ldots+a_n x^n$
$p(\sqrt{2}+\sqrt{3}+\sqrt{6})=0$
$\therefore \quad \quad x=\sqrt{2}+\sqrt{3}+\sqrt{6}$
$\Rightarrow \quad(x-\sqrt{6})^2=(\sqrt{2}+\sqrt{3})^2$
$\Rightarrow x^2-2 \sqrt{6} x+6=2+3+2 \sqrt{6}$
$\Rightarrow \quad\left(x^2+1\right)^2=2 \sqrt{6}(x+1)^2$
$\Rightarrow \quad x^4+2 x^2+1=24\left(x^2+2 x+1\right)$
$\Rightarrow x^4-22 x^2-48 x-23=0$
$\therefore p(x)=x^4-22 x^2-48 x-23=0$
$\text { Minimum value of } x=4$
Standard 11
Mathematics
