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

A quadratic equation is a polynomial equation of degree 2 in one variable — meaning the highest power of the variable appearing in it is 2. General form: ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. (The condition a ≠ 0 is essential — if a were 0, the x² term would vanish entirely and the equation would become linear, not quadratic.) A quadratic equation has at most two roots (solutions), which may be real and distinct, real and equal, or complex (not real at all), depending on the equation's discriminant.

For a quadratic equation ax² + bx + c = 0, the roots are given by the quadratic formula: x = [−b ± √(b² − 4ac)] / 2a The quantity under the square root, D = b² − 4ac, is called the discriminant. Case 1: D > 0. The square root of a positive number is a real, non-zero value, so the formula gives TWO DISTINCT real roots (one from +√D, one from −√D). Case 2: D = 0. The square root of zero is zero, so both the "+" and "−" branches of the formula give the SAME value: x = −b/2a. The equation has exactly one real root, counted as a repeated (equal) root. Case 3: D < 0. The square root of a negative number is not a real number, so the formula produces NO real roots — the equation's roots are complex (involving imaginary numbers). This is why simply computing D = b² − 4ac, without solving the full equation, immediately tells you how many real roots to expect and whether they're equal.

We need two numbers that multiply to give the constant term (10) and add to give the coefficient of x (7): these numbers are 2 and 5. x² + 7x + 10 = 0 x² + 2x + 5x + 10 = 0 x(x + 2) + 5(x + 2) = 0 (x + 2)(x + 5) = 0 Setting each factor to zero: x + 2 = 0 → x = −2 x + 5 = 0 → x = −5 Verification: (−2)² + 7(−2) + 10 = 4 − 14 + 10 = 0 ✓, (−5)² + 7(−5) + 10 = 25 − 35 + 10 = 0 ✓ The roots are x = −2 and x = −5.

Using x = [−b ± √(b² − 4ac)] / 2a, with a = 3, b = −8, c = 4. Discriminant: b² − 4ac = (−8)² − 4(3)(4) = 64 − 48 = 16 √16 = 4 x = [8 ± 4] / 6 x = 12/6 = 2, or x = 4/6 = 2/3 Verification: 3(2)² − 8(2) + 4 = 12 − 16 + 4 = 0 ✓ 3(2/3)² − 8(2/3) + 4 = 3(4/9) − 16/3 + 4 = 4/3 − 16/3 + 12/3 = 0 ✓ The roots are x = 2 and x = 2/3.

Equal roots occur exactly when the discriminant is zero: b² − 4ac = 0. Given: a = 3, b = −4, c = k. (−4)² − 4(3)(k) = 0 16 − 12k = 0 12k = 16 k = 16/12 = 4/3 For k = 4/3, the equation 3x² − 4x + 4/3 = 0 has equal (repeated) roots.

Let the width be w metres. Since the length is 3 m more than the width, the length is (w + 3) metres. Using Area = length × width: w(w + 3) = 54 w² + 3w − 54 = 0 Factorising: we need two numbers multiplying to −54 and adding to 3 — these are 9 and −6. (w + 9)(w − 6) = 0 w = −9 or w = 6 Since a width cannot be negative, we take w = 6 m. Length = w + 3 = 6 + 3 = 9 m Verification: 6 × 9 = 54 m² ✓ The rectangle's width is 6 m and its length is 9 m.

To determine if an equation has real roots, examine its discriminant: D = b² − 4ac. Given: a = 1, b = 3, c = 7. D = 3² − 4(1)(7) = 9 − 28 = −19 Since D = −19 is negative, the square root of D is not a real number. Because the quadratic formula requires taking the square root of the discriminant, a negative discriminant means the formula cannot produce a real result — therefore x² + 3x + 7 = 0 has no real roots.

Vieta's Formulas relate a quadratic equation's coefficients directly to the sum and product of its roots, without needing to solve for the roots individually: Sum of roots: p + q = −b/a Product of roots: p × q = c/a These relationships work in both directions — given the coefficients, the sum and product of the roots can be found immediately, and conversely, given a desired sum and product, the corresponding quadratic equation can be constructed directly as x² − (sum)x + (product) = 0.

By Vieta's Formulas, for 2x² + kx + 24 = 0, the product of the roots equals c/a = 24/2 = 12. Given one root is 4, find the other root (call it r): 4 × r = 12 r = 3 The other root is 3. To find k, use the sum of roots: sum = −b/a = −k/2 4 + 3 = −k/2 7 = −k/2 k = −14 Verification: substituting x = 4 into 2x² + kx + 24 with k = −14: 2(16) + (−14)(4) + 24 = 32 − 56 + 24 = 0 ✓ The other root is 3, and k = −14.