Question

The number of real solutions of the equation ${\left(\frac{9}{10}\right)}^x=-3+x-x^2$ is

• A. none
• B. more than two
• C. one
• D. two

Answer 1

The correct option is A none

${\left(\frac{9}{10}\right)}^x=-3+x-x^2\Rightarrow {\left(\frac{9}{10}\right)}^x=-\{{(x-\frac{1}{2})}^2+\frac{11}{4}\}$

$\because$ LHS is always positive and RHS is always negative for all $x$.

Hence, no solution.