Deeksha Vedantu Logo
Deeksha Vedantu Logo

Introduction to Quadratic Equations

Introduction to Quadratic Equations

Quadratic equations are a crucial part of algebra, forming the foundation for solving complex problems across various disciplines. Let’s dive into what quadratic equations are, explore their methods of solution, and review real-life applications.

What is a Quadratic Equation?

A quadratic equation is any equation that can be expressed as:

\boldsymbol{ax^2 + bx + c = 0}

Where:

  • \boldsymbol{a, b, c} are constants (\boldsymbol{a \neq 0}).
  • \boldsymbol{x} is the variable.

Standard Form of Quadratic Equations

The standard form of a quadratic equation is:

\boldsymbol{ax^2 + bx + c = 0}

Key components:

  • \boldsymbol{ax^2}: The quadratic term.
  • \boldsymbol{bx}: The linear term.
  • \boldsymbol{c}: The constant term.

Example:

Convert \boldsymbol{3x - 2x^2 + 5 = 0} into standard form:
Rearrange terms: \boldsymbol{-2x^2 + 3x + 5 = 0}.
Multiply by \boldsymbol{-1}: \boldsymbol{2x^2 - 3x - 5 = 0}.

Methods to Solve Quadratic Equations

1. Factorization

This involves expressing the quadratic equation as a product of two binomials.

Example: Solve \boldsymbol{x^2 - 7x + 10 = 0}.
Steps:

  1. Factorize: \boldsymbol{(x - 5)(x - 2) = 0}.
  2. Solve: \boldsymbol{x = 5}, \boldsymbol{x = 2}.

Answer: \boldsymbol{x = 5, x = 2}.

2. Completing the Square

This technique transforms the quadratic equation into a perfect square trinomial.

Example: Solve \boldsymbol{x^2 + 6x + 5 = 0}.
Steps:

  1. Rewrite: \boldsymbol{x^2 + 6x = -5}.
  2. Add \boldsymbol{(6/2)^2 = 9}: \boldsymbol{x^2 + 6x + 9 = 4}.
  3. Factorize: \boldsymbol{(x + 3)^2 = 4}.
  4. Solve: \boldsymbol{x + 3 = \pm 2}.
    • \boldsymbol{x = -1}, \boldsymbol{x = -5}.

Answer: \boldsymbol{x = -1, x = -5}.

3. Quadratic Formula

The quadratic formula is:

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

Example: Solve \boldsymbol{2x^2 + 3x - 2 = 0}.
Steps:

  1. Identify \boldsymbol{a = 2, b = 3, c = -2}.
  2. Discriminant: \boldsymbol{b^2 - 4ac = 3^2 - 4(2)(-2) = 9 + 16 = 25}.
  3. Apply formula:
    \displaystyle\boldsymbol{x = \frac{-3 \pm \sqrt{25}}{2(2)}}
    \displaystyle\boldsymbol{x = \frac{6 \pm 12}{4}}.

Solutions: \displaystyle\boldsymbol{x = \frac{1}{2}, x = -2}.

4. Graphical Method

In this method, the equation \boldsymbol{y = ax^2 + bx + c} is plotted as a parabola. The roots are the points where the parabola intersects the x-axis.

Example: For \boldsymbol{y = x^2 - 4x + 3}:

  1. Plot \boldsymbol{y = (x - 1)(x - 3)}.
  2. Intersection points: \boldsymbol{x = 1, x = 3}.

Answer: \boldsymbol{x = 1, x = 3}.

Nature of Roots

The discriminant (\boldsymbol{\Delta = b^2 - 4ac}) determines the type of roots:

  • \boldsymbol{\Delta > 0}: Two distinct real roots.
  • \boldsymbol{\Delta = 0}: One repeated root.
  • \boldsymbol{\Delta < 0}: Complex roots.

Example: Solve \boldsymbol{x^2 + 4x + 8 = 0}.

  1. Discriminant: \boldsymbol{\Delta = b^2 - 4ac = 16 - 32 = -16}.
  2. Complex roots:
    \displaystyle\boldsymbol{x = \frac{-4 \pm \sqrt{-16}}{2}}
    \boldsymbol{x = -2 + 2i, x = -2 - 2i}.

Answer: \boldsymbol{x = -2 + 2i, x = -2 - 2i}.

Applications of Quadratic Equations

  1. Projectile Motion: Calculating the path of an object in free fall.
  2. Geometry: Finding dimensions or areas of geometric shapes.
  3. Optimization: Maximizing profits or minimizing costs in economics.

Example:
A rectangle’s area is 24 m². If the length is 2 m more than the width, find the dimensions.

  1. Let width = \boldsymbol{x}. Length = \boldsymbol{x + 2}.
  2. Area: \boldsymbol{x(x + 2) = 24}.
  3. Solve \boldsymbol{x^2 + 2x - 24 = 0}: \boldsymbol{(x - 4)(x + 6) = 0}.
  4. \boldsymbol{x = 4} (positive value).

Answer: Width = 4 m, Length = 6 m.

Sample Questions with Answers

  1. Solve \boldsymbol{x^2 - 5x + 6 = 0}.
    Answer: \boldsymbol{x = 2, x = 3}.
  2. Find the nature of roots for \boldsymbol{x^2 + 4x + 4 = 0}.
    Answer: \boldsymbol{\Delta = 0}, one repeated root (\boldsymbol{x = -2}).
  3. Solve using the quadratic formula: \boldsymbol{x^2 - 3x - 10 = 0}.
    Answer: \boldsymbol{x = 5, x = -2}.

FAQs

Where are quadratic equations used?2024-12-18T11:08:52+05:30

Quadratic equations are used in physics, geometry, economics, engineering, and optimization problems.

What determines the nature of roots?2024-12-18T11:08:29+05:30

The discriminant (\boldsymbol{\Delta = b^2 - 4ac}) determines if roots are real, repeated, or complex.

What is the quadratic formula?2024-12-18T11:07:00+05:30

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

What are the methods to solve quadratic equations?2024-12-18T11:04:14+05:30

Factorization, completing the square, quadratic formula, and graphical method.

What is a quadratic equation?2024-12-18T11:02:40+05:30

A polynomial equation of degree two, expressed as \boldsymbol{ax^2 + bx + c = 0}, where \boldsymbol{a \neq 0}.

Related Topics

Join Deeksha Vedantu

> PU + Competitive Exam CoachingPreferred Choice For Toppers25+ Years of Academic Excellence70k+ Success Stories

Related Pages

Latest Posts

  • Choosing the Right Electives CBSE vs. ICSE Class 10
  • 10th Grade Physical Education Staying Fit and Active
  • Cultivating Leadership Skills in Teenagers
  • How to effectively Communicate with Your High School Teen
  • Difference Between IIT and NIT
  • Difference Between BITSAT and JEE

Contact Us

    By submitting my data, I authorize Deeksha and its representatives to Call, SMS, Email or WhatsApp me about its products and offers. This consent overrides any registration for DNC / NDNC., I agree to be contacted.

    Head Office

    Ace Creative Learning Pvt Ltd
    Deeksha House,
    163/B, 6th Main, 3rd Cross,
    JP Nagar 3rd Phase, Bengaluru,
    Karnataka – 560078