Question

For each equation given below, find the value (s) of $p$ so that the equation has equal roots:

$2p{x}^2-20x+\left(13p-1\right)=0$

Answer 1

Step 1: Solve for discriminant.

The given equation is $2p{x}^2-20x+\left(13p-1\right)=0$.

Compare the given equation with the standard quadratic equation $a{x}^2+bx+c=0$ to get:

$a=2p,b=-20$ and $c=13p-1$

Now, compute the discriminant as follows:

$$\begin{gathered} D=b^2-4ac \\ ={\left(-20\right)}^2-4\times 2p\times \left(13p-1\right) \\ =400-8p\left(13p-1\right) \\ =400-104p^2+8p \end{gathered}$$

Step 2: Apply the condition for equal roots.

Since the roots are equal, therefore $D=0$.

$$\begin{gathered} \begin{array}{rcl}& \Rightarrow & b^2-4ac=0\end{array} \\ \Rightarrow 400-104p^2+8p=0 \\ \Rightarrow 13p^2-p-50=0 \end{gathered}$$

The quadratic formula used to solve the quadratic equation $a{x}^2+bx+c=0$ is given as $\begin{array}{rcl}x& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{array}$.

Substitute $a=13,b=-1$ and $c=-50$ in the quadratic formula and simplify as follows:

$$\begin{gathered} \Rightarrow p=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\times 13\left(-50\right)}}{2\times 13} \\ \Rightarrow p=\frac{-\left(-1\right)\pm \hspace{0.17em}51}{2\times 13} \\ \Rightarrow p=\frac{1\pm \hspace{0.17em}51}{26} \\ \Rightarrow p=2,-\frac{25}{13} \end{gathered}$$

Thus, for $p=2$ or $p=-\frac{25}{13}$, the equation has equal roots.