Question

Given $f(x)=\frac{x^4-7x^2+9}{x-(3/x)+1}$. 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)

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