#### Question

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

\[kx\left( x - 2\sqrt{5} \right) + 10 = 0\]

#### Solution

The given quadratic equation is \[kx\left( x - 2\sqrt{5} \right) + 10 = 0\] and roots are real and equal.

Then find the value of *k.*

Here,

\[kx\left( x - 2\sqrt{5} \right) + 10 = 0\]

\[ \Rightarrow k x^2 - 2\sqrt{5}kx + 10 = 0\]

So,\[a = k, b = - 2\sqrt{5}k \text { and c }= 10 .\]

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

Putting the value of

\[a = k, b = - 2\sqrt{5}k \text { and } c = 10 .\]

\[D = \left( - 2\sqrt{5}k \right)^2 - 4\left( k \right)\left( 10 \right)\]

\[ = 20 k^2 - 40k\]

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

So, \[20 k^2 - 40k = 0\]

Now factorizing the above equation,

\[20 k^2 - 40k = 0\]

\[ \Rightarrow 20k\left( k - 2 \right) = 0\]

\[ \Rightarrow 20k = 0 \text { or } k - 2 = 0\]

\[ \Rightarrow k = 0 \text { or } k = 2\]

Therefore, the value of \[k = 0, 2\].