Question

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

The correct option is C $16$

We have $f\left(x\right)=\frac{x^4-7x^2+9}{x-\left(3/x\right)+1}$

$f\left(x\right)$ can be written as

$f\left(x\right)=\frac{x(x^4-7x^2+9)}{x^2-3+x}$

zeroes of the $x^4-7x^2+9=0$

$x^2=\frac{7\pm \sqrt{49-36}}{2}=\frac{7\pm \sqrt{13}}{2}$

$x=\sqrt{\frac{7\pm \sqrt{13}}{2}}$

multiply numerator and denominator by $\sqrt{2}$, we get

$x=\sqrt{\frac{14\pm 2\sqrt{13}}{4}}$

$x=\frac{\sqrt{14\pm 2\sqrt{13}}}{2}$

$x=\frac{\sqrt{13+1\pm 2\sqrt{13}}}{2}$

$x=\frac{\sqrt{(1\pm \sqrt{13}{)}^2}}{2}$

$x=\frac{(1\pm \sqrt{13})}{2}$

$\therefore a+b+c=13+1+2=16$