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 $3x^2+mx+2=0\dots$
The quadratic equation has equal roots if its discriminant is zero
$$\begin{array}{cc}& \Rightarrow D=b^2-4ac=0\\ & \Rightarrow m^2-4\left(2\right)\left(3\right)=0\\ & \Rightarrow m^2=24\\ & \Rightarrow m=\pm 2\sqrt{6}\end{array}$$
When $m=2\sqrt{6}$, equation (i) becomes
$$\begin{array}{cc}& 3x^2+2\sqrt{6}x+2=0\\ & \Rightarrow (\sqrt{3}x+\sqrt{2}{)}^2=0\\ & \Rightarrow x=-\frac{\sqrt{2}}{\sqrt{3}}=-\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=-\frac{\sqrt{6}}{3}\end{array}$$
When $m=-2\sqrt{6}$, equation (i) becomes
$$\begin{array}{cc}& 3x^2-2\sqrt{6}x+2=0\\ & \Rightarrow (\sqrt{3}x-\sqrt{2}{)}^2=0\\ & \Rightarrow x=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\\ & -\frac{\sqrt{6}}{3},\frac{\sqrt{6}}{3}\end{array}$$