Quadratic Equations - Previous Year Questions Solved Examples (Class 10 Mathematics)
Quadratic equations are solved using various methods including factorization, completing the square, and the quadratic formula. These questions test concep
TL;DR: Quadratic equations are solved using various methods including factorization, completing the square, and the quadratic formula. These questions test c…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
Quadratic equations are solved using various methods including factorization, completing the square, and the quadratic formula. These questions test concep
Quadratic Equations - Previous Year Questions — Solved Numerical Examples (Step by Step)
Example 1: Solve the equation 2x² - 5x + 3 = 0 by factorization. [3 marks]
Solution: 2x² - 5x + 3 = 0. Find two numbers that multiply to give 2 × 3 = 6 and add to give -5. These are -2 and -3. Rewrite: 2x² - 2x - 3x + 3 = 0. Factor: 2x(x - 1) - 3(x - 1) = 0. (2x - 3)(x - 1) = 0. Therefore x = 3/2 or x = 1.
Example 2: For the equation ax² + bx + c = 0, the sum of roots is -b/a and product of roots is c/a. If the roots of a quadratic equation are 3 and -5, form the equation. [2 marks]
Solution: Sum of roots = 3 + (-5) = -2. Product of roots = 3 × (-5) = -15. Using x² - (sum)x + (product) = 0: x² - (-2)x + (-15) = 0. x² + 2x - 15 = 0.
Example 3: Solve 4x² - 4x + 1 = 0 using the quadratic formula. [3 marks]
Solution: For ax² + bx + c = 0, x = (-b ± √(b² - 4ac)) / 2a. Here a = 4, b = -4, c = 1. Discriminant = (-4)² - 4(4)(1) = 16 - 16 = 0. x = (4 ± 0) / 8 = 4/8 = 1/2. Since discriminant = 0, there is one repeated root.
Example 4: If x² - 7x + 12 = 0, find the nature of roots without solving. [2 marks]
Solution: Discriminant = b² - 4ac = (-7)² - 4(1)(12) = 49 - 48 = 1 > 0. Since discriminant is positive, the equation has two distinct real roots.
Example 5: A number is 27 more than the square of its opposite. Find the number. [3 marks]
Solution: Let the number be x. Its opposite is -x. According to the problem: x = (-x)² + 27. x = x² + 27. x² - x + 27 = 0. Discriminant = 1 - 108 = -107 < 0. No real solution. This problem has no real solution.
Example 6: A rectangular field has area 1200 m². If length is 20 m more than breadth, find the dimensions. [3 marks]
Solution: Let breadth = x m. Length = (x + 20) m. Area = length × breadth. x(x + 20) = 1200. x² + 20x - 1200 = 0. By quadratic formula or factorization: x² + 40x - 20x - 1200 = 0. x(x + 40) - 20(x + 40) = 0. (x - 20)(x + 40) = 0. x = 20 (since x > 0). Breadth = 20 m, Length = 40 m.
Example 7: Prove that the equation 2x² - 4x + 3 = 0 has no real roots. [2 marks]
Solution: Discriminant = b² - 4ac = (-4)² - 4(2)(3) = 16 - 24 = -8 < 0. Since discriminant is negative, the equation has no real roots.
Tips
- Always check the discriminant first to determine nature of roots before attempting to solve.
- For word problems, define the variable clearly and set up the equation carefully.
- Factorization works only when roots are rational; use quadratic formula for irrational roots.
- Verify solutions by substituting back into the original equation.
Frequently Asked Questions
What does discriminant tell us?
If Δ > 0: two distinct real roots. If Δ = 0: one repeated real root. If Δ < 0: no real roots.
When should I use completing the square method?
Use completing the square to derive the quadratic formula or when the equation is difficult to factorize.
More Mathematics Solved Examples
- Real Numbers (HCF LCM Euclid)
- Pair of Linear Equations in Two Variables
- Circles
- Statistics
- Linear Equations in Two Variables
- Heron's Formula
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