#### Question

If the equation 9x^{2} + 6kx + 4 = 0 has equal roots, then the roots are both equal to

\[\pm \frac{2}{3}\]

\[\pm \frac{3}{2}\]

0

±3

#### Solution

The given quadric equation is 9x^{2} + 6kx + 4 = 0, and roots are equal.

Then find roots of given equation*.*

Here, a = 9, b = 6k and , c = 4

As we know that `D = b^2 - 4ac`

Putting the value of a = 9, b = 6k and , c = 4

` = (6k)^2 - 4 xx 9 xx 4`

` = 36k^2 - 144`

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

`36k^2 - 144 = 0`

`36(k^2 - 4) = 0`

`k^2 - 4 = 0`

` k^2 = 4`

`k = ±2`

So, putting the value of* k *in quadratic equation

When k = 2then equation be and when k = -2 then

`9x^2 + 6xx 2x + 4 = 0`

`9x^2 + 12x + 4 = 0`

`9x^2 + 6x xx 6x + 4 = 0`

`3x (3x + 2) + 2 (3x + 2) = 0`

and

`9x^2 + 6x - 2x + 4 = 0`

`9x^2 - 12x + 4 = 0`

`9x^2 - 6x - 6x + 4 = 0`

`3x (3x - 2) - 2 (3x - 2) = 0`

Therefore, the value of `x = ± 2/3`.