Quadratic Equations
CBSE 2026 Board Exam Set 1 (Code 30/7/1)
CBSE_BOARD_PYQ_2026_30_7_1
Grade 10
Question:
[Section B]
Find the nature of the roots of the quadratic equation $2x^2 - 6x + 3 = 0$. If real roots exist, find them.
Step-by-Step Solution
Key Concept: $D = b^2 - 4ac = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0$. Roots are $\dfrac{-b \pm \sqrt{D}}{2a}$.
$D = (-6)^2 - 4(2)(3) = 36 - 24 = 12 > 0$. Since $D > 0$, roots are real and distinct. [1.0 Mark]
$x = \dfrac{6 \pm \sqrt{12}}{4} = \dfrac{6 \pm 2\sqrt{3}}{4} = \dfrac{3 \pm \sqrt{3}}{2}$. [1.0 Mark]
Correct Answer: Real and distinct roots: $\dfrac{3 \pm \sqrt{3}}{2}$
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