Question

Find the value of $p$ so that the equation $3x^2-5x+2p=0$ has equal roots. Also, find the roots.

Answer 1

Step 1:

Comparison of the coefficients:

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

$\begin{array}{rcl}a& =& 3\\ b& =& -5\\ c& =& 2p\end{array}$

Step 2:

Find the discriminant:

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

$\begin{array}{rcl}D& =& b^2-4ac\\ & =& {\left(-5\right)}^2-\left(4\times 3\times 2p\right)\\ & =& 25-24p\end{array}$

Since, the given equation has equal roots, so the discriminant must be zero.

$$\begin{gathered} D=0 \\ \begin{array}{rcl}& \Rightarrow & 25-24p=0\\ & \Rightarrow & 24p=25\\ & \Rightarrow & p=\frac{25}{24}\end{array} \end{gathered}$$

Step 3:

Find the roots:

Put the value of $p$ in given equation.

$\begin{array}{rcl}3x^2-5x+\left(2\times \frac{25}{24}\right)& =& 0\\ & \Rightarrow & 3x^2-5x+\frac{25}{12}=0\end{array}$

Use the quadratic formula to solve the equation.

$\begin{array}{rcl}x& =& \frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =& \frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-\left(4\times 3\times {\displaystyle \frac{25}{12}}\right)}}{2\times 3}\\ & =& \frac{5\pm \sqrt{25-25}}{6}\\ & =& \frac{5\pm 0}{6}\\ & =& \frac{5}{6}\end{array}$

Hence, the value of $p=\frac{25}{24}$ and each root is $\frac{5}{6}$.