Question

The values of k for which the quadratic equation  \[16 x^2 + 4kx + 9 = 0\]  has real and equal roots are

 

पर्याय

• \[6, - \frac{1}{6}\]

• 36, −36

•  6, −6

• \[\frac{3}{4}, - \frac{3}{4}\]

Answer 1

The given quadratic equation  \[16 x^2 + 4kx + 9 = 0\]

has equal roots.

Here, 

\[a = 16, b = 4k \text { and } c = 9\] .

As we know that 

\[D = b^2 - 4ac\]

Putting the values of  \[a = 16, b = 4k \text { and } c = 9\].

\[D = \left( 4k \right)^2 - 4\left( 16 \right)\left( 9 \right)\]

\[ = 16 k^2 - 576\]

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

Thus, 

\[16 k^2 - 576 = 0\]

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

\[ \Rightarrow (k + 6)(k - 6) = 0\]

\[ \Rightarrow k + 6 = 0 \text { or } k - 6 = 0\]

\[ \Rightarrow k = - 6 \text { or } k = 6\]

Therefore, the value of k is 6, −6.