Question

Let $f:R\to R$ be defined as $f\left(x\right)=x^2+5x+9$ then $f^{-1}\left(8\right)$ equals to

• A. $\{\frac{-5+\sqrt{20}}{2},\frac{-5-\sqrt{21}}{2}\}$
• B. $\{\frac{-5+\sqrt{21}}{2},\frac{-5-\sqrt{21}}{2}\}$
• C. $\{\frac{5-\sqrt{21}}{2},\frac{21-\sqrt{5}}{2}\}$
• D. Does not exist.

Answer 1

The correct option is B $\{\frac{-5+\sqrt{21}}{2},\frac{-5-\sqrt{21}}{2}\}$

Let $f^{-1}\left(8\right)=x,$ then $f\left(x\right)=8$

$⟹x^2+5x+9=8$

$⟹x^2+5x+1=0$

$\Rightarrow x=\frac{-5\pm \sqrt{25-4}}{2}$

$x=\frac{-5+\sqrt{21}}{2},\frac{-5-\sqrt{21}}{2}\in R$

Hence
$f^{-1}\left(8\right)=\{\frac{-5+\sqrt{21}}{2},\frac{-5-\sqrt{21}}{2}\}$