#### Question

Find the values of *k* for which the roots are real and equal in each of the following equation:\[px(x - 3) + 9 = 0\]

#### Solution

The given quadratic equation is \[px(x - 3) + 9 = 0\] and roots are real and equal.

Then find the value of p.

Here,

\[px(x - 3) + 9 = 0\]

\[ \Rightarrow p x^2 - 3px + 9 = 0\]

So,

\[a = p, b = - 3p \text { and } c = 9 .\]

As we know that \[D = b^2 - 4ac\]

Putting the value of \[a = p, b = - 3p \text { and } c = 9 .\]

\[D = \left( - 3p \right)^2 - 4\left( p \right)\left( 9 \right)\]

\[ = 9 p^2 - 36p\]

The given equation will have real and equal roots, if D = 0.

So,

\[9 p^2 - 36p = 0\]

Now factorizing the above equation,

\[9 p^2 - 36p = 0\]

\[ \Rightarrow 9p\left( p - 4 \right) = 0\]

\[ \Rightarrow 9p = 0 \text { or } p - 4 = 0\]

\[ \Rightarrow p = 0 \text { or } p = 4\]

Therefore, the value of \[p = 0, 4 .\]