#### Question

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

\[6, - \frac{1}{6}\]

36, −36

6, −6

\[\frac{3}{4}, - \frac{3}{4}\]

#### Solution

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.

Is there an error in this question or solution?

#### APPEARS IN

Solution The Values of K for Which the Quadratic Equation 16 X 2 + 4 K X + 9 = 0 Has Real and Equal Roots Are Concept: Solutions of Quadratic Equations by Factorization.