Question

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

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

Answer 1

Answer: (A)

Let $f(x)=-3+x-x^2$
Then, $f(x)<0$ for all $x$ because coefficient of $x^2<0$ and disc $<0$.
Thus, LHS of the given equation is always positive whereas the RHS is always less than zero. Hence, the given equation has no solution. Alternate Given equation is
$\frac{9}{10}=-3+x-x^2$

Let $y=\frac{9}{10}$, therefore
$y=-3+x-x^2$
$y=-\left[x^2-x+\frac{1}{4}\right]-3+\frac{1}{4}$
$\Rightarrow y+\frac{11}{4}=-{\left(x-\frac{1}{2}\right)}^2$
It is clear from the graph that two curves do not intersect. Hence, no solution exist.