Question

If $a^2-4a+1=4$, then the value of $\frac{a^3-a^2+a-1}{a^2-1}(a^2\ne 1)$ is equal to

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

The correct option is A $2$

Given,

$a^2-4a+1=4$

$\Rightarrow a^2-4a-3=0$

upon solving the above quadratic equation, we get,

$a=2\pm \sqrt{7}$

Given,

$\frac{a^3-a^2+a-1}{a^2-1}$

$=\frac{(2\pm \sqrt{7}{)}^3-(2\pm \sqrt{7}{)}^2+(2\pm \sqrt{7})-1}{(2\pm \sqrt{7}{)}^2-1}$

upon solving the above equation, we get,

$=\frac{40+16\sqrt{7}}{10+4\sqrt{7}}$

$=\frac{8(5+2\sqrt{7})}{2(5+2\sqrt{7})}$

$=2$