Question

Find the values of k for which the roots are real and equal in the following equation:

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

In the following, determine the values of k for which the given quadratic equation has equal roots:

kx(x – 3) + 9 = 0

Answer 1

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

Then find the value of k.

Here,

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

\[ \Rightarrow kx^2 - 3kx + 9 = 0\]

So,

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

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

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

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

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

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

So,

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

Now factorizing the above equation,

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

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

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

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

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