Question

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

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 $m=\frac{2}{3}$ and $n=\frac{3}{2}$, therefore,

The sum of the roots is:

$m+n=\frac{2}{3}+\frac{3}{2}=\frac{(2\times 2)+(3\times 3)}{3\times 2}=\frac{4+9}{6}=\frac{13}{6}$

And the product of the roots is:

$mn=\frac{2}{3}\times \frac{3}{2}=1$

Therefore, the required quadratic equation is

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

$\Rightarrow x^2-\frac{13}{6}x+1=0\Rightarrow 6x^2-13x+6=0$

Hence, $6x^2-13x+6=0$ is the quadratic equation whose roots are $\frac{2}{3}$ and $\frac{3}{2}$.