Question

Form the quadratic equation whose roots are $-3,\frac{3}{2}$

Answer 1

We know that if $m$ and $n$ are the roots of a quadratic equation $a{x}^2+bx+c=0$, then the sum of the roots is $(m+n)$ and the product of the roots is $\left(mn\right)$. And then the quadratic equation becomes $x^2-(m+n)x+mn=0$

Here, it is given that the roots of the quadratic equation are $m=-3$ and $n=\frac{3}{2}$, therefore,

The sum of the roots is:

$m+n=-3+\frac{3}{2}=\frac{-6+3}{2}=-\frac{3}{2}$

And the product of the roots is:

$mn=-3\times \frac{3}{2}=-\frac{9}{2}$

Therefore, the required quadratic equation is

$x^2-(m+n)x+mn=0$

$\Rightarrow x^2-(-\frac{3}{2})x+(-\frac{9}{2})=0\Rightarrow x^2+\frac{3}{2}x-\frac{9}{2}=0$

Hence, $x^2+\frac{3}{2}x-\frac{9}{2}=0$ is the quadratic equation whose roots are $3$ and $\frac{3}{2}$.