Question

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

$6x^2+12x-p=0$

Answer 1

Step 1: Solve for discriminant.

The given equation is as follows:

$6x^2+12x-p=0$

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

$a=6,b=12,c=-p$

Now, discriminant can be calculated as:

, $\begin{array}{rcl}& \Rightarrow & D=b^2-4ac\\ & =& {\left(12\right)}^2-4\times 6\times \left(-p\right)\\ & =& 144+24p\end{array}$

Step 2: Apply the condition for equal roots.

For equal roots,

$\begin{array}{rcl}{b}^2-4ac& =& 0\\ 144+24p& =& 0\\ 144& =& -24p\\ p& =& -\frac{144}{24}\\ p& =& -6\end{array}$

Thus for $p=-6$, the equation has equal roots.