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Q. 8 Let $f(x)=\frac{x^4-7 x^2+9}{x-\frac{3}{x}+1}$. If the zeros of the function $f(x)$ are of the form $\frac{a \pm \sqrt{b}}{c}$ where $a, b$ and $c$ are positive integers then the $\operatorname{sum}( a + b + c )$ is divisible by

Relations and Functions - Part 2

Solution:

$f(x)=\frac{x\left[\left(x^2-3\right)^2-x^2\right]}{\left[x^2+x-3\right]}=x\left(x^2-x-3\right)$
If $f ( x )=0 \Rightarrow x =0$ (rejected)
$x=\frac{1 \pm \sqrt{13}}{2} \equiv \frac{ a \pm \sqrt{ b }}{ c } $
$\therefore( a + b + c )=16 \Rightarrow 16 \text { is divisible by } 1,2,4,8$