Question

The value of k for which the equation $kx^2 - 6x - 4 = 0$ has real and equal roots, is

• A. $\frac{9}{4}$
• B. $-4$
• C. $-\frac{9}{4}$
• D. $-2$

Hint:

For any quadratic equation $ax^2 + bx + c = 0$ to have real and equal roots, remember the direct condition $b^2 = 4ac$.
This allows you to quickly set up the equation and solve for the unknown parameter without extra intermediate steps.

Answer 1

The Correct Option is C

Step 1: Use the fact that equal roots repeat the same value.
For real and equal roots, both roots equal $-\frac{b}{2a}$. Here $a = k$, $b = -6$, so each root is $\frac{6}{2k} = \frac{3}{k}$.
Step 2: Use the product of the roots instead of the discriminant.
Since both roots equal $\frac{3}{k}$, their product is $\left(\frac{3}{k}\right)^2 = \frac{9}{k^2}$. But the product of roots also equals $\frac{c}{a} = \frac{-4}{k}$.
Step 3: Equate the two expressions for the product and solve for k.
$\frac{9}{k^2} = \frac{-4}{k} \implies 9 = -4k \implies k = -\frac{9}{4}$.
\[ \boxed{k = -\frac{9}{4}} \]