Question

Given $f(x)=\frac{x^4-7 x^2+9}{x-(3 / x)+1}$. Its zeroes are of the form $\frac{a \pm \sqrt{b}}{c}$, where $a, b$ and $c$ are positive integers. Then the value of $(a+b+c)$, is

• A. 14
• B. 15
• C. 16
• D. 17

Answer 1

Answer: (C)

$f(x)=\frac{x^2\left(x^2+\frac{9}{x^2}-7\right)}{\left(x-\frac{3}{x}+1\right)}=\frac{x^2\left(\left(x-\frac{3}{x}\right)^2-1\right)}{\left(x-\frac{3}{x}+1\right)}=x^2\left(x-\frac{3}{x}-1\right)$
$f ( x )=0, \text { gives } x ^2- x -3=0 $
$ x =\frac{1 \pm \sqrt{1+12}}{2}=\frac{1 \pm \sqrt{13}}{2}$
$\therefore a =1, b =13, c =2$
$\therefore a + b + c =16$