Question

Find the value of ‘$p$’ for which the roots of the following equations are real and equal:

$x^2-2\left(p+1\right)x+p^2=0$

Answer 1

Step 1:

Comparison of the coefficients:

Compare the coefficients of the given equation with the standard quadratic equation, $a{x}^2+bx+c=0$.

$\begin{array}{rcl}a& =& 1\\ b& =& -2\left(p+1\right)\\ c& =& p^2\end{array}$

Step 2:

Determine the value of $p$.

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

$\begin{array}{rcl}D& =& b^2-4ac\\ & =& {\left(-2\left(p+1\right)\right)}^2-\left(4\times 1\times p^2\right)\\ & =& {\left(2p+2\right)}^2-4p^2\\ & =& 4p^2+4+8p-4p^2\\ & =& 8p+4\end{array}$

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

$$\begin{gathered} D=0 \\ \begin{array}{rcl}& \Rightarrow & 8p+4=0\\ & \Rightarrow & p=-\frac{4}{8}\\ \therefore p& =& -\frac{1}{2}\end{array} \end{gathered}$$

Hence, the required value is $p=-\frac{1}{2}$.