Question

Find the value of $c$ such that equation $4x^2-2\left(c+1\right)x+\left(c+4\right)=0$ has real and equal roots.

Answer 1

Step 1:

Comparison of the coefficients:

Compare the coefficients of the given equation with the standard equation, $A{x}^2+Bx+C=0$.

$$\begin{gathered} A=4 \\ B=-2\left(c+1\right) \\ C=c+4 \end{gathered}$$

Step 2:

Find the value of $c$:

The discriminant $D$ of the given equation can be calculated as,

$\begin{array}{rcl}D& =& B^2-4AC\\ & =& {\left(-2\left(c+1\right)\right)}^2-\left(4\times 4\times \left(c+4\right)\right)\\ & =& {\left(2c+2\right)}^2-16c-64\\ & =& 4c^2+8c+4-16c-64\\ & =& 4c^2-8c-60\\ & =& 4(c^2-2c-15)\end{array}$

For real and equal roots, the discriminant must be zero.

$$\begin{gathered} D=0 \\ \Rightarrow 4(c^2-2c-15)=0 \\ \Rightarrow c^2-2c-15=0 \\ \begin{array}{rcl}& \Rightarrow & c^2-5c+3c-15=0\\ & \Rightarrow & c\left(c-5\right)+3\left(c-5\right)=0\\ & \Rightarrow & \left(c+3\right)\left(c-5\right)=0\\ & \Rightarrow & c+3=0\mathrm{or}c-5=0\\ & \Rightarrow & c=-3\mathrm{or}c=5\end{array} \end{gathered}$$

Hence, $c=-3\mathrm{and}c=5$ are the required values of $c$.