Question

Find the number of real solution to the quadratic equation
$-(x+2)(x-1)=3$

Answer 1

Solve the quadratic equation $-(x+2)(x-1)=3$ as follows:

$-(x+2)(x-1)=3$

$\Rightarrow -(x^2-x+2x-2)=3$

$\Rightarrow -(x^2+x-2)=3$

$\Rightarrow -x^2-x-1=0$

$\Rightarrow x^2+x+1=0$

Finding the roots of the above quadratic equation $x^2+x+1=0$ as shown below:

$x=\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1\pm \sqrt{-3}}{2}=\frac{-1\pm \sqrt{3i}}{2}$

The quadratic equation has imaginary roots and hence, it has no real roots.