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Quadratic Equations — Telangana (SSC) Class 10 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Quadratic Equations" — important questions with detailed answers, download PDF for board exam preparation.

TL;DR: Free step-by-step Telangana (SSC) Class 10 Mathematics solutions for "Quadratic Equations" — important questions with detailed answers, download PDF f…

By Syllab.in · Updated Jun 14, 2026

Q1: Define a quadratic equation and state the general form.

Quadratic Equation: A quadratic equation in one variable is a polynomial equation of degree 2. General form: ax² + bx + c = 0 where a, b, and c are real numbers and a ≠ 0. The condition a ≠ 0 is essential because if a = 0, it becomes a linear equation, not quadratic. Examples of quadratic equations: - x² + 5x + 6 = 0 - 2x² - 7x + 3 = 0 - x² - 9 = 0 - 3x² + 2x = 0 Characteristics: - The highest power of the variable is 2 - A quadratic equation has at most 2 real roots - The graph of y = ax² …

Q2: Prove that the discriminant determines the nature of roots of a quadratic equation.

For the quadratic equation ax² + bx + c = 0, the discriminant is Δ = b² - 4ac. The nature of roots is determined as follows: 1) If Δ > 0 (discriminant is positive): - The equation has two distinct real roots - Roots are given by x = (-b ± √Δ)/(2a) - Example: x² + 5x + 6 = 0 has Δ = 25 - 24 = 1 > 0, so two distinct real roots 2) If Δ = 0 (discriminant is zero): - The equation has two equal (repeated) real roots - Both roots are x = -b/(2a) - Example: x² - 4x + 4 = 0 has…

Q3: Solve x² + 5x + 6 = 0 by factorisation.

Given equation: x² + 5x + 6 = 0 Step 1: Identify factors We need two numbers that multiply to 6 and add to 5. Factors of 6 are: 1×6, 2×3 Checking: 2 + 3 = 5 ✓ and 2 × 3 = 6 ✓ Step 2: Factorise x² + 5x + 6 = (x + 2)(x + 3) = 0 Step 3: Apply zero product property Either (x + 2) = 0 or (x + 3) = 0 Therefore: x = -2 or x = -3 Step 4: Verification For x = -2: (-2)² + 5(-2) + 6 = 4 - 10 + 6 = 0 ✓ For x = -3: (-3)² + 5(-3) + 6 = 9 - 15 + 6 = 0 ✓ Therefore, the roots are x = -2 and x = -3

Q4: Solve 2x² - 7x + 3 = 0 using the quadratic formula.

Given equation: 2x² - 7x + 3 = 0 Identify coefficients: a = 2, b = -7, c = 3 Step 1: Calculate the discriminant Δ = b² - 4ac = (-7)² - 4(2)(3) = 49 - 24 = 25 Since Δ = 25 > 0, the equation has two distinct real roots. Step 2: Apply the quadratic formula x = (-b ± √Δ)/(2a) x = (-(-7) ± √25)/(2 × 2) x = (7 ± 5)/4 Step 3: Calculate the roots Root 1: x = (7 + 5)/4 = 12/4 = 3 Root 2: x = (7 - 5)/4 = 2/4 = 1/2 Step 4: Verification For x = 3: 2(3)² - 7(3) + 3 = 18 - 21 + 3 = 0 ✓ For x = …

Q5: For what values of k does the equation 2x² - 3x + k = 0 have equal roots?

For a quadratic equation to have equal roots, the discriminant must be zero: Δ = 0 Given equation: 2x² - 3x + k = 0 Coefficients: a = 2, b = -3, c = k Step 1: Write the discriminant condition Δ = b² - 4ac = 0 (-3)² - 4(2)(k) = 0 9 - 8k = 0 Step 2: Solve for k 8k = 9 k = 9/8 Step 3: Verification With k = 9/8, the equation becomes: 2x² - 3x + 9/8 = 0 Multiplying by 8: 16x² - 24x + 9 = 0 Discriminant: (-24)² - 4(16)(9) = 576 - 576 = 0 ✓ The equal root is: x = -b/(2a) = -(-3)/(2 × 2) = 3/4 Ve…

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