Introduction to Quadratic Equations and Standard Form for CBSE Class 10

In earlier classes, you studied linear equations, where the highest exponent of the variable is $1$. While linear relationships describe uniform growth and straight lines, many real-world phenomena—such as the trajectory of a kicked football, the area of a plot of land, or the acceleration of an automobile—involve quantities multiplied by themselves. This gives rise to second-degree polynomial equations, known as quadratic equations.

In CBSE Class 10 Mathematics, Chapter 4 (Quadratic Equations) is one of the most scoring and foundational units in algebra. Mastering standard form and learning how to set up quadratic models from real-world descriptions is the critical first step in solving higher-level board examination problems.

What You Will Learn

• Formal definition of a quadratic equation in one variable
• The standard form: $ax^2 + bx + c = 0$ and the mandatory $a \ne 0$ condition
• How to test and simplify algebraic expressions to determine whether they are quadratic
• Formulating quadratic equations from geometric and practical situations
• Meaning of the roots (solutions) of a quadratic equation
• Step-by-step solved CBSE board examination questions and common pitfalls

1. What is a Quadratic Equation?

A quadratic equation in the variable $x$ is an algebraic equation of the second degree.

Standard Form of a Quadratic Equation

Any equation that can be written in the form: $ax^2 + bx + c = 0$ where $a, b, c$ are real numbers and $a \ne 0$, is called a quadratic equation in standard form.

Here:

• $a$ is the coefficient of $x^2$ (the leading coefficient).
• $b$ is the coefficient of $x$.
• $c$ is the constant term.
• $x$ is the unknown variable.

Important: <u>The coefficient of $x^2$ must never be zero in a quadratic equation ($a \ne 0$). If $a = 0$, the quadratic term $ax^2$ vanishes, reducing the equation to $bx + c = 0$, which is a linear equation, not a quadratic equation!</u>

Examples of Quadratic Equations:

• $2x^2 + x - 300 = 0$ (Standard form with $a = 2, b = 1, c = -300$)
• $x^2 - 49 = 0$ (Here $b = 0$, giving $a = 1, b = 0, c = -49$)
• $3x^2 - 5x = 0$ (Here $c = 0$, giving $a = 3, b = -5, c = 0$)

2. Testing Whether an Equation is Quadratic

An equation may not appear to be quadratic at first glance, or it might deceptively appear quadratic when it is not. You must always simplify the equation completely by expanding brackets and moving all terms to one side before determining its degree.

                           Testing for Quadratic Equations
                                          |
                +-------------------------+-------------------------+
                |                                                   |
    Expand & Combine Like Terms                             Inspect Highest Power
                |                                                   |
   Bring to form: ax² + bx + c = 0              If power = 2 and a ≠ 0  → Quadratic!
                                                If power ≠ 2 or a = 0  → NOT Quadratic!

Solved Example 1: Checking Quadratic Nature

Problem: Check whether the following equations are quadratic:

1. $(x - 2)^2 + 1 = 2x - 3$
2. $x(x + 1) + 8 = (x + 2)(x - 2)$
3. $(x + 2)^3 = x^3 - 4$

Solution:

Part 1: $(x - 2)^2 + 1 = 2x - 3$

1. Expand the LHS using $(a - b)^2 = a^2 - 2ab + b^2$: $x^2 - 4x + 4 + 1 = 2x - 3$ $x^2 - 4x + 5 = 2x - 3$
2. Transpose all terms to LHS: $x^2 - 4x - 2x + 5 + 3 = 0 \implies x^2 - 6x + 8 = 0$
3. This is in the form $ax^2 + bx + c = 0$ with $a = 1 \ne 0, b = -6, c = 8$.
4. Therefore, <u>it is a quadratic equation</u>.

Part 2: $x(x + 1) + 8 = (x + 2)(x - 2)$

1. Expand LHS and RHS: $x^2 + x + 8 = x^2 - 4$
2. Transpose all terms to LHS: $x^2 - x^2 + x + 8 + 4 = 0 \implies x + 12 = 0$
3. Here, the $x^2$ term cancels out ($a = 0$). The highest power of $x$ is $1$.
4. Therefore, <u>it is NOT a quadratic equation</u> (it is a linear equation).

Part 3: $(x + 2)^3 = x^3 - 4$

1. Expand LHS using $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$: $x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 = x^3 - 4$ $x^3 + 6x^2 + 12x + 8 = x^3 - 4$
2. Subtract $x^3$ from both sides: $6x^2 + 12x + 8 + 4 = 0 \implies 6x^2 + 12x + 12 = 0$
3. Dividing by 6 gives $x^2 + 2x + 2 = 0$.
4. Even though a cubic term was originally visible, it cancelled out, leaving an equation of degree $2$.
5. Therefore, <u>it is a quadratic equation</u>.

3. Formulating Quadratic Equations from Situations

A vital skill tested in board exams is translating word descriptions into standard quadratic equations.

Solved Example 2: Area of a Rectangular Plot

Problem: The area of a rectangular plot is $528\text{ m}^2$. The length of the plot (in metres) is one more than twice its breadth. Represent this situation in the form of a quadratic equation.

Solution:

1. Let the breadth of the rectangular plot be $x\text{ metres}$.
2. According to the problem, the length is one more than twice the breadth: $\text{Length} = (2x + 1)\text{ metres}$
3. We know that: $\text{Area of Rectangle} = \text{Length} \times \text{Breadth}$ $(2x + 1) \times x = 528$ $2x^2 + x = 528$
4. Rearranging in standard form ($= 0$): $2x^2 + x - 528 = 0$
5. This is the required quadratic equation representing the situation.

4. Roots of a Quadratic Equation

A real number $\alpha$ is called a root (or solution) of the quadratic equation $ax^2 + bx + c = 0$ if substituting $x = \alpha$ satisfies the equation: $a\alpha^2 + b\alpha + c = 0$

Remember: If $\alpha$ is a root of $ax^2 + bx + c = 0$, then $(x - \alpha)$ is a factor of the quadratic polynomial $P(x) = ax^2 + bx + c$.

5. Summary and Examination Tips

Exam Tip: In questions asking to "Represent the following situation in the form of a quadratic equation", do NOT solve for $x$ unless specifically asked. Simply write the equation in standard form $ax^2 + bx + c = 0$ with correct physical units declared for variables.

Common Mistake: Forgetting that an equation must have an equals sign ($=$). Writing $ax^2 + bx + c$ is a quadratic polynomial, whereas writing $ax^2 + bx + c = 0$ is a quadratic equation!