Question

If $f\left(x\right)=\frac{x^2+x-6}{x^2-6x+8}$, solve $f\left(x\right)=3$.

• A. $\{-5,-1\}$
• B. $\{2,7.5\}$
• C. $\{\frac{1+3\sqrt{7}}{2},\frac{1-3\sqrt{7}}{2}\}$
• D. $\{\frac{17+\sqrt{73}}{6},\frac{17-\sqrt{73}}{6}\}$
• E. $\oslash$

Answer 1

The correct option is B $\{2,7.5\}$

$f\left(x\right)=\frac{x^2+x-6}{x^2-6x+8}$

If $f\left(x\right)=3$, then

$\Rightarrow 3=\frac{x^2+x-6}{x^2-6x+8}$

$\Rightarrow 3x^2-18x+24=x^2+x-6$

$\Rightarrow 2x^2-19x+30=0$

$\Rightarrow 2x^2-15x-4x+30=0$

$\Rightarrow x(2x-15)-2(2x-15)=0$

$\Rightarrow (2x-15)(x-2)=0$

$\Rightarrow x=2,\frac{15}{2}=2,7.5$