Free Grade 10 Quadratic Equations Study Material
Comprehensive free study resources for grade 10 quadratic equations covering all solution methods, nature of roots, and practical applications. Perfect for CBSE, ICSE, and state board students preparing for board exams and competitive tests.
2.3 Quadratic Equations
- Standard form of quadratic equation (ax² + bx + c = 0)
- Solution methods: factorization and completing the square
- Understanding discriminant and nature of roots
- Quadratic formula derivation and application
- Real-world applications of quadratic equations
Quadratic Equations
Grade 10 Mathematics – CBSE/ICSE/NCERT Curriculum
Introduction to Quadratic Equations
A quadratic equation is a polynomial equation of degree 2 in one variable \( x \), expressed in the standard form:
\[ ax^2 + bx + c = 0, \quad a \ne 0 \]
where:
- \( a, b, c \) are real numbers,
- \( a \) is the coefficient of \( x^2 \),
- \( b \) is the coefficient of \( x \),
- \( c \) is the constant term.
Examples
- \( 2x^2 + 3x – 5 = 0 \) (Quadratic)
- \( x^2 – 4 = 0 \) (Quadratic)
- \( 3x + 7 = 0 \) (Not quadratic, degree 1)
Factorization Method
Steps:
- Write the equation in standard form: \( ax^2 + bx + c = 0 \).
- Factorize the quadratic expression into two linear factors.
- Set each factor equal to zero and solve for \( x \).
Example 1: Solve \( x^2 – 5x + 6 = 0 \).
Solution:
\[ \begin{align*} x^2 – 5x + 6 &= 0 \\ x^2 – 3x – 2x + 6 &= 0 \\ x(x – 3) – 2(x – 3) &= 0 \\ (x – 3)(x – 2) &= 0 \\ \Rightarrow x = 3 &\quad \text{or} \quad x = 2 \end{align*} \]Example 2: Solve \( x^2 + 7x + 12 = 0 \)
Solution:
\[ \begin{align*} x^2 + 7x + 12 &= 0 \\ x^2 + 3x + 4x + 12 &= 0 \\ x(x + 3) + 4(x + 3) &= 0 \\ (x + 4)(x + 3) &= 0 \\ x = -4 &\quad \text{or} \quad x = -3 \end{align*} \]Exercise: Solve the following using factorization method
- \( x^2 + 11x + 24 = 0 \)
- \( x^2 – 4x – 21 = 0 \)
- \( 2x^2 – 5x – 3 = 0 \)
Completing the Square Method
Concept: This method converts a quadratic expression into a perfect square.
Steps:
- Make the coefficient of \( x^2 \) equal to 1 (if needed).
- Bring the constant term to the right-hand side.
- Add \( \left( \frac{b}{2} \right)^2 \) to both sides.
- Write the left side as a perfect square.
- Solve for \( x \) by taking square roots.
Example: Solve \( x^2 + 6x + 5 = 0 \)
Solution:
\[ \begin{align*} x^2 + 6x &= -5 \\ x^2 + 6x + 9 &= -5 + 9 \\ (x + 3)^2 &= 4 \\ x + 3 &= \pm 2 \\ x &= -1 \quad \text{or} \quad x = -5 \end{align*} \]Exercise: Solve using the Completing the Square Method
- \( x^2 + 10x + 16 = 0 \)
- \( x^2 – 6x + 5 = 0 \)
Quadratic Formula
The quadratic formula solves any quadratic equation of the form:
\[ ax^2 + bx + c = 0 \]The roots are given by:
\[ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \]Example: Solve \( 2x^2 + 5x – 3 = 0 \)
Solution:
\[ \begin{align*} x &= \frac{-5 \pm \sqrt{25 + 24}}{4} \\ &= \frac{-5 \pm 7}{4} \\ x &= 0.5 \quad \text{or} \quad x = -3 \end{align*} \]Exercise: Solve using the Quadratic Formula
- \( x^2 + 7x + 10 = 0 \)
- \( 2x^2 – 3x + 1 = 0 \)