Question

Find the value of ‘p’ for which the given quadratic equation has equal roots $p{k^2} + 12k + 9 = 0$

Answer 1

Hint: In this question we have to find the value of p if the given equation has equal roots.
If in a quadratic equation if roots are real and distinct then \[{b^2} - 4ac > 0\],
If in a quadratic equation if the roots are real and equal then \[{b^2} - 4ac = 0\] and
If in a quadratic equation if the roots are imaginary then \[{b^2} - 4ac < 0\].
Here, we have given the roots are equal and we can find the value of p by applying the above formula in accordance with the given condition.

Complete step-by-step answer:
We are given with a quadratic equation i.e. $p{k^2} + 12k + 9 = 0$ and the roots of this equation are equal. We have to find the value of p.

As we know that the roots of the equation can be obtained for a quadratic equation $a{x^2} + bx + c = 0$ by the formula given below:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, by comparing it with given quadratic equation i.e. $p{k^2} + 12k + 9 = 0$ we have,
$a = p$, $b = 12$ and $c = 9$ ……..(1)

If the roots are equal in the given quadratic equation then
\[ \Rightarrow {b^2} - 4ac = 0\]
By putting the values of a, b and c from (1), we get,
\[ \Rightarrow {12^2} - 4(p)(9) = 0\]
As we know the square of 12 is 144 and put this value in above equation, we get,
\[ \Rightarrow 144 - 4(p)(9) = 0\]
By opening the brackets, we get,
\[ \Rightarrow 144 - 36p = 0\]
By taking the p term on one side and constant other we get,
\[ \Rightarrow - 36p = - 144\]
By cancelling negative side of both sides, we get,
\[ \Rightarrow 36p = 144\]
\[ \Rightarrow p = \dfrac{{144}}{{36}}\]
By dividing numerator with denominator, we get,
\[ \Rightarrow p = 4\]
Therefore, \[p = 4\] is the required answer.

Note: We may solve this question by obtaining both the roots and then equate them if students forget the above approach. We know the roots of a quadratic equation are given by:
$\begin{gathered}
   \Rightarrow \alpha = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} \\
   \Rightarrow \beta = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} \\
\end{gathered} $
If both the roots are equal then
$ \Rightarrow \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
Cancelling 2a from both sides, we get,
$ \Rightarrow - b + \sqrt {{b^2} - 4ac} = - b - \sqrt {{b^2} - 4ac} $
Now cancel -b from both sides, we get,
$ \Rightarrow \sqrt {{b^2} - 4ac} = - \sqrt {{b^2} - 4ac} $
$\begin{gathered}
   \Rightarrow 2\sqrt {{b^2} - 4ac} = 0 \\
   \Rightarrow \sqrt {{b^2} - 4ac} = 0 \\
\end{gathered} $
By squaring both sides, we get,
$ \Rightarrow {b^2} - 4ac = 0$
As you can see that the formula we used in the solution is the same as above. Now, we can find the value of p.