Question

Find the values of m for which equation $3x^2+mx+2=0$ has equal roots. Also, find the roots of the given equation.

Answer 1

Given: Quadratic equation is $3x^2+mx+2=0$

We know that, for equal roots of $a{x}^2+bx+c=0$, its discriminant$\left(D\right)$ is zero

Discriminant$\left(D\right)=b^2-4ac=0$

Here, $a=3,b=m,c=2$

$\Rightarrow D=m^2-4\times 3\times 2=0$

$\Rightarrow m^2=24$

$\Rightarrow m=\pm 2\sqrt{6}$

Putting $m=2\sqrt{6}$ in the given equation, it becomes

$3x^2+2\sqrt{6}x+2=0$

$\Rightarrow (\sqrt{3}x+\sqrt{2}{)}^2=0$

$\{\because a^2+2ab+b^2=(a+b{)}^2\}$

$\Rightarrow x=\frac{-\sqrt{2}}{\sqrt{3}}=\frac{-\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{-\sqrt{6}}{3}$

When $m=-2\sqrt{6}$, equation becomes $3x^2-2\sqrt{6}x+2=0$

$\Rightarrow (\sqrt{3}x-\sqrt{2}{)}^2=0$

$\{\because a^2-2ab+b^2=(a-b{)}^2\}$

$\Rightarrow x=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}$

$\therefore$ For $m=2\sqrt{6},x=\frac{-\sqrt{6}}{3}$ and for $m=-2\sqrt{6},x=\frac{\sqrt{6}}{3}$