Question

The value of k, for which the equation $2x^{2}kx+3=0$ has two real equal roots, are

(a) $\pm 2\sqrt{3}$
(b) $\pm 3\sqrt{2}$
(c) $\pm 2\sqrt{6}$
(d) $\pm \sqrt{6}$

Answer 1

(c) $\pm 2\sqrt{6}$
The given equation is 2x² + kx + 3 = 0.
This is of the form ax² + bx + c = 0, where, a = 2, b = k and c = 3.
Therefore,
D = (b² - 4ac) = (k² - 4 × 2 × 3) = (k² - 24)
For real and equal roots, we must have:
D = 0 ⇒ k² - 24 = 0 ⇒k² = 24 ⇒ $k=\pm \sqrt{24}=\pm 2\sqrt{6}$
Hence, $2\sqrt{6}$ and $-2\sqrt{6}$ are the required values of k.