Find the value of ‘ $p$ ’ for which the roots of the following equations are real and equal:
$9 x^{2} - 24 x + p = 0$

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Solution

Step 1:
Comparison of the coefficients:
Compare the coefficients of the given equation with the standard quadratic equation, $a x^{2} + b x + c = 0$ .
$\begin{array}{rcl} a & = & 9 \\ b & = & - 24 \\ c & = & p \end{array}$
Step 2:
Find the discriminant:
The discriminant $D$ of the given equation can be calculated as,
$\begin{array}{rcl} D & = & b^{2} - 4 a c \\ = & {(- 24)}^{2} - (4 \times 9 \times p) \\ = & 576 - 36 p \end{array}$
Therefore, the discriminant is $D = 576 - 36 p$ .
For real and equal roots, the discriminant must be 0 $(D = 0)$ .
$\begin{array}{rcl} 576 - 36 p & = & 0 \\ \Rightarrow & 36 p = 576 \\ \Rightarrow & p = \frac{576}{36} \\ \Rightarrow & p = 16 \end{array}$
Hence, the value of $p$ is 16.

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