Chapter – 4

Quadratic Equations

In this post we have given the detailed notes of class 10 Maths Chapter 4 (Quadratic Equations) in English. These notes are useful for the students who are going to appear in class 10 board exams.

Chapter 4: Quadratic Equations

What is a Quadratic Equation?

An equation in which the highest power of the variable is 2 is called a quadratic equation. It is also called an “equation of degree 2”.

Any equation of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0, is called a quadratic equation in the variable x. Here a is the coefficient of x², b is the coefficient of x, and c is the constant term.

Roots of a Quadratic Equation

A real number α is called a root of the quadratic equation ax² + bx + c = 0 if aα² + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the equation.

The zeroes of the quadratic polynomial ax² + bx + c and the roots of the quadratic equation ax² + bx + c = 0 are the same thing.

A quadratic equation can have at most two real roots.

Worked Example — Finding Roots by Factorisation

Consider the equation 2x² − 3x + 1 = 0. Let us find its roots by splitting the middle term.

2x² − 3x + 1 = 0
⇒ 2x² − 2x − x + 1 = 0
⇒ 2x(x − 1) − 1(x − 1) = 0
⇒ (x − 1)(2x − 1) = 0

So x = 1 or x = ½.

Hence x = 1, ½ are the roots of the equation 2x² − 3x + 1 = 0.

Methods of Solving a Quadratic Equation

There are three standard methods used to solve quadratic equations:

1. Factorisation Method
2. Completing the Square Method
3. Quadratic Formula Method

1. Factorisation Method

In this method, the middle term (bx) is split into two parts such that their sum is equal to the middle term, and the product of the two parts is equal to the product of the first term (ax²) and the last term (c).

Worked Example — Word Problem Solved by Factorisation

Question: John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. Find out how many marbles they had to start with.

Solution: Let the number of marbles John had be x. Then the number of marbles Jivanti had = 45 − x.

After losing 5 marbles each:
Marbles with John = x − 5
Marbles with Jivanti = 45 − x − 5 = 40 − x

According to the question:
(x − 5)(40 − x) = 124
⇒ 40x − x² − 200 + 5x = 124
⇒ −x² + 45x − 200 = 124
⇒ x² − 45x + 324 = 0

This is a quadratic equation in x. Splitting the middle term:
x² − 36x − 9x + 324 = 0
⇒ x(x − 36) − 9(x − 36) = 0
⇒ (x − 36)(x − 9) = 0

So x = 36 or x = 9.

If John had 36 marbles, Jivanti had 45 − 36 = 9 marbles. If John had 9 marbles, Jivanti had 45 − 9 = 36 marbles. (Both answers are valid, depending on which friend is which.)

Worked Example — Finding a Root by Substitution

Consider the quadratic equation 2x² − 3x + 1 = 0. If we substitute x = 1 on the left-hand side, we get: (2 × 1²) − (3 × 1) + 1 = 0, which equals the right-hand side. So x = 1 is a root of this equation, and 1 is a zero of the quadratic polynomial 2x² − 3x + 1.

Another Worked Example — Factorisation

Question: Find the roots of 2x² − 5x + 3 = 0 by factorisation.

Solution: Split the middle term −5x as −2x − 3x, since (−2x) × (−3x) = 6x² = (2x²) × 3.

2x² − 5x + 3 = 2x² − 2x − 3x + 3
= 2x(x − 1) − 3(x − 1)
= (2x − 3)(x − 1)

So the roots are found by setting each factor equal to zero:
2x − 3 = 0 ⇒ x = 3/2
x − 1 = 0 ⇒ x = 1

Hence x = 3/2 and x = 1 are the roots of the given equation. In other words, 3/2 and 1 are the zeroes of the polynomial 2x² − 5x + 3.

Note: We obtain the roots of the quadratic equation by factorising it into two linear factors and equating each factor to zero.

2. Completing the Square Method

When a quadratic equation of the form ax² + bx + c is converted into the form a(x + h)² + k, this process is called completing the square.

Worked Example — Word Problem Solved by Completing the Square

Question: The product of Sunita’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age?

Solution: Let her present age be x years. Then her age two years ago = (x − 2), and her age four years from now = (x + 4).

According to the question:
(x − 2)(x + 4) = 2x + 1
⇒ x² + 2x − 8 = 2x + 1
⇒ x² − 9 = 0

So Sunita’s present age satisfies the quadratic equation x² − 9 = 0.

We can write this as x² = 9. Taking square roots, x = 3 or x = −3.

Since age cannot be negative, x = 3. So Sunita’s present age is 3 years.

Steps for Completing the Square

Consider the equation x² + 4x − 5 = 0. We convert it into the form (x + a)² − b² = 0, from which the roots can be found easily.

x² + 4x = x² + 2(x)(2)
= x² + 2(x)(2) + 2² − 2²
= (x + 2)² − 4

So x² + 4x − 5 = (x + 2)² − 4 − 5 = (x + 2)² − 9.

Thus x² + 4x − 5 = 0 can be written as (x + 2)² − 9 = 0, i.e. (x + 2)² = 9, giving x + 2 = ±3, so x = 1 or x = −5.

Worked Example — Completing the Square with a Fractional Coefficient

Consider the equation 3x² − 5x + 2 = 0. Dividing throughout by 3:
x² − (5/3)x + 2/3 = 0

Now, {x − (5/6)}² − (5/6)² + 2/3
= (x − 5/6)² + 2/3 − 25/36
= (x − 5/6)² − 1/36
= (x − 5/6)² − (1/6)²

So (x − 5/6) = ±1/6, which gives x = 1 or x = 2/3.

Worked Example — Word Problem

Question: The sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution: Let the side of the first square be x m and the side of the second square be y m.

Sum of areas: x² + y² = 468 … (1)
Difference of perimeters: 4x − 4y = 24, i.e. x − y = 6 … (2)

From (2), x = 6 + y. Substituting in (1):
(6 + y)² + y² = 468
⇒ 36 + 12y + 2y² = 468
⇒ y² + 6y − 216 = 0

Completing the square: (y + 3)² = 225, so y + 3 = ±15, giving y = 12 or y = −18.

Since the side of a square cannot be negative, y = 12. Substituting back in (2), x = 18.

So the sides of the two squares are 18 m and 12 m.

Quadratic Formula (Sridharacharya’s Formula)

By completing the square on the general equation ax² + bx + c = 0, the roots of any quadratic equation are given by the formula:

x = [−b ± √(b² − 4ac)] / 2a, provided b² − 4ac ≥ 0.

Nature of Roots — The Discriminant

In the quadratic formula, the expression b² − 4ac is called the discriminant of the quadratic equation, usually denoted by D, because it “discriminates” between the possible natures of the roots.

For a quadratic equation ax² + bx + c = 0, a ≠ 0:

• If b² − 4ac > 0, the equation has two distinct real roots: x = [−b + √(b² − 4ac)] / 2a and x = [−b − √(b² − 4ac)] / 2a.
• If b² − 4ac = 0, the equation has two equal real roots (coincident roots), each equal to −b/2a.
• If b² − 4ac < 0, there is no real number whose square is b² − 4ac, so the equation has no real roots.

Worked Example — Nature of Roots

Question: Find the discriminant of the quadratic equation 2x² − 4x + 3 = 0, and hence find the nature of its roots.

Solution: Comparing with ax² + bx + c = 0, we get a = 2, b = −4, c = 3.

Discriminant D = b² − 4ac = (−4)² − 4(2)(3) = 16 − 24 = −8 < 0.

Since D < 0, the given equation has no real roots.

Worked Example — Checking if a Situation is Possible

Question: Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth. (Illustrative example based on the discriminant test.)

Solution: Let the breadth of the park be x m. Since the perimeter is 80 m, length + breadth = 40 m, so length = (40 − x) m.

Area = x(40 − x) = 400
⇒ 40x − x² = 400
⇒ x² − 40x + 400 = 0

Discriminant D = (−40)² − 4(1)(400) = 1600 − 1600 = 0.

Since D = 0, the equation has two equal real roots, so this situation is possible, and the length and breadth of the park are both equal to 20 m (i.e. it must be a square).

Worked Example — Two Friends’ Ages (No Real Solution)

Question: Is the following situation possible? If so, find their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages (in years) was 48.

Solution: Let the present age of one friend be x years. Then the present age of the other = (20 − x) years.

Four years ago, their ages were (x − 4) and (16 − x). Product = 48:
(x − 4)(16 − x) = 48
⇒ 16x − x² − 64 + 4x = 48
⇒ −x² + 20x − 64 = 48
⇒ x² − 20x + 112 = 0

Discriminant D = (−20)² − 4(1)(112) = 400 − 448 = −48 < 0.

Since D < 0, this equation has no real roots. So the given situation is not possible.

Key Points to Remember

• A quadratic equation in variable x is of the form ax² + bx + c = 0, where a, b, c are real numbers and a ≠ 0.
• A real number α is a root of ax² + bx + c = 0 if aα² + bα + c = 0. A quadratic equation has at most two real roots.
• Quadratic equations can be solved by three methods: factorisation, completing the square, and the quadratic formula.
• Quadratic Formula: x = [−b ± √(b² − 4ac)] / 2a, valid when b² − 4ac ≥ 0.
• The Discriminant is D = b² − 4ac. It decides the nature of the roots.
• If D > 0 → two distinct real roots.
• If D = 0 → two equal real roots.
• If D < 0 → no real roots.
• In factorisation, split the middle term so that the sum of the two parts equals the middle term and their product equals the product of the first and last terms.
• Always check that the value of the variable makes sense in the context of the word problem (e.g. age or length cannot be negative).

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