Question

Find the value of $m$ so that the roots of the equation:

$\left(4-m\right)x^2+\left(2m+4\right)x+\left(8m+1\right)=0$ are equal.

Answer 1

Step 1: Given information in the question.

The given quadratic equation is as follows:

$\left(4-m\right)x^2+\left(2m+4\right)x+\left(8m+1\right)=0$

On comparing it with a general quadratic equation, we get:

$a=4-m$, $b=2m+4$ and $c=8m+1$

Step 2: Apply the condition of equal roots

$\begin{array}{rcl}{b}^2-4ac& =& 0\\ {\left(2m+4\right)}^2-4\times \left(4-m\right)\times \left(8m+1\right)& =& 0\\ 4m^2+16+16m-\left(16-4m\right)\left(8m+1\right)& =& 0\\ 4m^2+16+16m-128m+32m^2-16+4m& =& 0\\ 36m^2-108m& =& 0\end{array}$

Step 3: Factorise the obtained quadratic equation.

$\begin{array}{rcl}36m^2-108m& =& 0\\ 36m(m-3)& =& 0\\ m& =& 0\text{or}3\end{array}$

Hence, the required values of $m$ are $0\text{and}3$.