Question

The number of real solutions of the equation $\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)=-3+x-x^2$ is

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is A

$0$

Explanation of correct answer

Given $\left(\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)=-3+x-x^2$

Let $f\left(x\right)=-6+x-x^2$

For a quadratic equation $a{x}^2+bx+c=0$ the roots are real if $b^2-4ac\ge 0$

Now comparing $f\left(x\right)$ with the standard form we get

$$\begin{gathered} \therefore b^2-4ac \\ =1-4\left(-6\right)\left(-1\right) \\ =1-24 \\ =-23 \end{gathered}$$

Therefore, the equation has no solution.

Hence, the correct option is $\left(\mathrm{A}\right)$.