Question

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

(a) $6,-\frac{1}{6}$
(b) 36, −36
(c) 6, −6
(d) $\frac{3}{4},-\frac{3}{4}$

Answer 1

The given quadratic equation $16x^2+4kx+9=0$, has equal roots.

Here, $a=16,b=4k\mathrm{and}c=9$.

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

Putting the values of $a=16,b=4k\mathrm{and}c=9$.

$$\begin{gathered} D={\left(4k\right)}^2-4\left(16\right)\left(9\right) \\ =16k^2-576 \end{gathered}$$

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

Thus, $16k^2-576=0$
$$\begin{gathered} \Rightarrow k^2-36=0 \\ \Rightarrow (k+6)(k-6)=0 \\ \Rightarrow k+6=0\mathrm{or}k-6=0 \\ \Rightarrow k=-6\mathrm{or}k=6 \end{gathered}$$

Therefore, the value of k is 6, −6.

Hence, the correct option is (c).