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Mathematics — CBSE Class 10 board question

Q1. [1]
The value of $k$ for which the equation $kx^2 - 6x - 4 = 0$ has real and equal roots, is
  1. (A) $\frac{9}{4}$
  2. (B) $-4$
  3. (C) $-\frac{9}{4}$
  4. (D) $-2$
Previously asked in CBSE board exam
2026 30/5/1 Q1
Generated by claude-sonnet-4-6 · 2026-06-15 10:33 · grounding rag
Model Answer

(C) $-\dfrac{9}{4}$

Here $a = k$, $b = -6$, $c = -4$. For equal roots, $b^2 - 4ac = 0$:
$$(-6)^2 - 4(k)(-4) = 0 \implies 36 + 16k = 0 \implies k = -\frac{9}{4}$$

Source: Chapter 4, Section 4.4 Nature of Roots

Explanation

For equal roots, the discriminant $b^2 - 4ac = 0$. Substituting the coefficients and solving gives $k = -\dfrac{9}{4}$. Note that $k \neq 0$ (otherwise it wouldn't be a quadratic equation), and this value satisfies that condition.

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