Quadratic equation class 10

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quadratic equation class 10

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Quadratic Equation Class 10

Answer:
Quadratic equations are an essential topic in Class 10 mathematics, forming a foundation for both higher-level math and various real-world applications. Here is a detailed overview of quadratic equations suited for a Class 10 syllabus.

Definition and Standard Form

A quadratic equation is a second-degree polynomial equation in a single variable x, with the general form:

ax^2 + bx + c = 0

where a, b, and c are constants, and a \neq 0.

Solution of Quadratic Equations

Quadratic equations can be solved using various methods, including:

  1. Factorization
  2. Completing the Square
  3. Quadratic Formula
  4. Graphical Method

1. Factorization

Factorization involves expressing the quadratic equation as a product of two binomials. Consider the quadratic equation:

x^2 + 5x + 6 = 0

Steps to Factorize:

  • Identify two numbers that multiply to give the constant term (6) and add to give the middle coefficient (5). In this case, the numbers are 2 and 3.
  • Express the equation in factorized form:
    (x + 2)(x + 3) = 0
  • Find the roots by setting each factor to zero:
    x + 2 = 0 \implies x = -2
    x + 3 = 0 \implies x = -3

Thus, the solutions are x = -2 and x = -3.

2. Completing the Square

This method involves rewriting the quadratic equation in the form of a perfect square trinomial. Consider the equation:

x^2 + 6x + 5 = 0

Steps to Complete the Square:

  • Move the constant to the other side:
    x^2 + 6x = -5
  • Add the square of half the coefficient of x to both sides:
    x^2 + 6x + 9 = 4
  • Write as a perfect square:
    (x + 3)^2 = 4
  • Solve for x:
    x + 3 = \pm 2
    x = -3 \pm 2
    x = -1 \quad \text{or} \quad x = -5

Thus, the solutions are x = -1 and x = -5.

3. Quadratic Formula

The general quadratic formula for solving ax^2 + bx + c = 0 is:

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Example:
Solve 2x^2 - 4x - 6 = 0:

  • Identify coefficients: a = 2, b = -4, c = -6.
  • Substitute into the formula:
    x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot (-6)}}{2 \cdot 2}
    x = \frac{4 \pm \sqrt{16 + 48}}{4}
    x = \frac{4 \pm \sqrt{64}}{4}
    x = \frac{4 \pm 8}{4}
    • Solutions are:
      x = 3 \quad \text{or} \quad x = -1

4. Graphical Method

Graphically, a quadratic equation corresponds to a parabola on the coordinate plane. The solutions (roots) of the quadratic equation are the x-coordinates where the parabola intersects the x-axis.

Key Points:

  • The parabola opens upwards if a > 0 and downwards if a < 0.
  • The vertex of the parabola is given by x = -\frac{b}{2a}.
  • The discriminant (b^2 - 4ac) determines the nature of the roots:
    • If \Delta > 0, there are two distinct real roots.
    • If \Delta = 0, there is one real root.
    • If \Delta < 0, there are no real roots, but two complex roots.

Practice Problems

  1. Solve 3x^2 - 12x + 9 = 0 by factorization.
  2. Find the roots of x^2 + 4x - 5 = 0 by completing the square.
  3. Use the quadratic formula to solve 2x^2 + 3x - 2 = 0.
  4. Determine the nature of the roots of the equation x^2 - 4x + 4 = 0.

Final Answer:
Understanding and mastering these methods for solving quadratic equations is crucial for students, providing them with tools to tackle a broad range of mathematical problems both in exams and real-life applications.

This detailed explanation aims to help students understand the core concepts and methods involved in solving quadratic equations covered in Class 10 mathematics. If there are specific problems or more advanced topics you’d like to discuss, feel free to ask!