Quadratic Equations Class 10 Mathematics — Revision Notes

This chapter explores quadratic equations of the form ax^2 + bx + c = 0. Students learn solution methods including factorization and quadratic formula, and

Quadratic Equations Class 10 Mathematics — Revision Notes

This chapter explores quadratic equations of the form ax^2 + bx + c = 0. Students learn solution methods including factorization and quadratic formula, and

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TL;DR: This chapter explores quadratic equations of the form ax^2 + bx + c = 0. Students learn solution methods including factorization and quadratic formula…

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This chapter explores quadratic equations of the form ax^2 + bx + c = 0. Students learn solution methods including factorization and quadratic formula, and

Quadratic Equation Basics

  • Standard form: ax^2 + bx + c = 0 where a ≠ 0
  • Degree is 2, having at most 2 real roots
  • If b = 0: ax^2 + c = 0 called pure quadratic
  • If c = 0: ax^2 + bx = 0 called incomplete quadratic

Solution by Factorization

  • Factorize ax^2 + bx + c into product of linear factors
  • Set each factor equal to zero and solve
  • Works when roots are rational and can be easily found
  • Example: x^2 - 5x + 6 = 0 becomes (x-2)(x-3) = 0, so x = 2 or 3

Quadratic Formula

  • For ax^2 + bx + c = 0: x = (-b ± sqrt(b^2 - 4ac)) / 2a
  • Discriminant D = b^2 - 4ac determines nature of roots
  • D > 0: two distinct real roots
  • D = 0: two equal real roots
  • D < 0: no real roots (two complex roots)

Applications

  • Speed and distance problems: set up quadratic from given conditions
  • Geometry problems: area, perimeter conditions lead to quadratic
  • Ages and work problems converted to quadratic equations
  • Check that roots are valid for given problem context

Key Terms

  • Quadratic Equation: Polynomial equation of degree 2 in the form ax^2 + bx + c = 0
  • Discriminant: Value b^2 - 4ac determining nature of roots of quadratic
  • Roots: Values of variable satisfying the quadratic equation

Frequently Asked Questions

When do quadratic equations have no real roots?

When discriminant D = b^2 - 4ac < 0. In this case, roots are complex (involve imaginary unit i).

How do you solve x^2 - 4x + 3 = 0?

By factorization: (x-1)(x-3) = 0, so x = 1 or 3. Or using formula: x = (4 ± sqrt(16-12))/2 = (4 ± 2)/2, giving x = 3 or 1.

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