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

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 + bx + c is a parabola

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 Δ = 16 - 16 = 0, so equal roots 3) If Δ < 0 (discriminant is negative): - The equation has no real roots - Roots are complex conjugates - Example: x² + x + 1 = 0 has Δ = 1 - 4 = -3 < 0, so no real roots This proves that the discriminant completely determines the nature of roots.

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

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 = 1/2: 2(1/2)² - 7(1/2) + 3 = 2(1/4) - 7/2 + 3 = 1/2 - 7/2 + 3 = 0 ✓ Therefore, the roots are x = 3 and x = 1/2

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 Verification: 2(3/4)² - 3(3/4) + 9/8 = 2(9/16) - 9/4 + 9/8 = 9/8 - 18/8 + 9/8 = 0 ✓ Therefore, k = 9/8

Let the width of the rectangle be x m. Then the length is (x + 2) m. Step 1: Form the equation using area formula Area = length × width 48 = (x + 2) × x 48 = x² + 2x x² + 2x - 48 = 0 Step 2: Solve using factorisation Find two numbers that multiply to -48 and add to 2. Factors: 8 and -6 (since 8 × (-6) = -48 and 8 + (-6) = 2) x² + 2x - 48 = (x + 8)(x - 6) = 0 Step 3: Find the roots Either x + 8 = 0 or x - 6 = 0 Therefore: x = -8 or x = 6 Since width cannot be negative, x = 6 m. Step 4: Find dimensions Width = 6 m Length = x + 2 = 6 + 2 = 8 m Step 5: Verification Area = 8 × 6 = 48 m² ✓ Length is 2 m more than width: 8 - 6 = 2 m ✓ Therefore, the dimensions are length = 8 m and width = 6 m

Given equation: x² + 2x + 5 = 0 Coefficients: a = 1, b = 2, c = 5 Step 1: Calculate the discriminant Δ = b² - 4ac = (2)² - 4(1)(5) = 4 - 20 = -16 Step 2: Interpret the result Since Δ = -16 < 0 (discriminant is negative), the equation has no real roots. This means: - The parabola y = x² + 2x + 5 does not intersect the x-axis - The equation has two complex conjugate roots (not real) Alternative verification using completing the square: x² + 2x + 5 = 0 (x² + 2x + 1) + 5 - 1 = 0 (x + 1)² + 4 = 0 (x + 1)² = -4 Since (x + 1)² is always ≥ 0 for real x, it cannot equal -4. Therefore, there are no real solutions. Conclusion: The equation x² + 2x + 5 = 0 has no real roots

Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. For a quadratic equation ax² + bx + c = 0 with roots p and q: 1) Sum of roots: p + q = -b/a 2) Product of roots: p × q = c/a Proof: The quadratic equation with roots p and q can be written as: a(x - p)(x - q) = 0 Expanding: a(x² - qx - px + pq) = 0 a(x² - (p + q)x + pq) = 0 ax² - a(p + q)x + apq = 0 Comparing with ax² + bx + c = 0: Coefficient of x: b = -a(p + q) → p + q = -b/a Constant term: c = apq → pq = c/a Example: For 2x² - 5x + 3 = 0 with roots p and q: p + q = -(-5)/2 = 5/2 pq = 3/2 We can verify: the roots are 1 and 3/2 1 + 3/2 = 5/2 ✓ 1 × 3/2 = 3/2 ✓ Applications: - Constructing polynomials when roots are known - Finding relationships between roots without solving explicitly - Checking if roots satisfy required conditions

Given equation: 3x² + kx + 15 = 0 One root: p = 5/3 Find: other root q and the value of k Step 1: Use Vieta's formulas Coefficients: a = 3, b = k, c = 15 Product of roots: pq = c/a (5/3) × q = 15/3 (5/3) × q = 5 q = 5 ÷ (5/3) q = 5 × (3/5) q = 3 Step 2: Find k using sum of roots Sum of roots: p + q = -b/a (5/3) + 3 = -k/3 5/3 + 9/3 = -k/3 14/3 = -k/3 k = -14 Step 3: Verification With k = -14, the equation is: 3x² - 14x + 15 = 0 Check p = 5/3: 3(5/3)² - 14(5/3) + 15 = 3(25/9) - 70/3 + 15 = 25/3 - 70/3 + 45/3 = (25 - 70 + 45)/3 = 0/3 = 0 ✓ Check q = 3: 3(3)² - 14(3) + 15 = 27 - 42 + 15 = 0 ✓ Therefore, the other root is 3 and k = -14