ICSE Class 8 Maths Chapter 18 Quadratic Equations PDF Download

Chapter 18: Quadratic Equations

Quadratic Equation

An equation of the form \(ax^2 + bx + c = 0\), where a, b and c are real numbers and a ≠ 0, is called a quadratic equation in the variable x.

For Example:

\(x^2 - 3x + 2 = 0\) and \(6x^2 + x - 1 = 0\) are quadratic equations in the variable x.

A number α is a root (or solution) of the quadratic equation \(ax^2 + bx + c = 0\) if it satisfies the equation i.e. if \(aα^2 + bα + c = 0\).

For Example:

When we substitute x = 2 in the quadratic equation \(x^2 - 3x + 2 = 0\), we get \(2^2 - 3 \times 2 + 2 = 0\) i.e. \(4 - 6 + 2 = 0\) i.e. 0 = 0, which is true. Therefore, 2 is a root of the quadratic equation \(x^2 - 3x + 2 = 0\).

When we substitute x = 3 in the quadratic equation \(x^2 - 3x + 2 = 0\), we get \(3^2 - 3 \times 3 + 2 = 0\) i.e. \(9 - 9 + 2 = 0\) i.e. 2 = 0, which is wrong. Therefore, 3 is not a root of the quadratic equation \(x^2 - 3x + 2 = 0\).

Solving Quadratic Equations

Factorisation can be used to solve quadratic equations. The equation \(x^2 - 3x + 2 = 0\) can be written as (x - 1)(x - 2) = 0. This equation can be solved by using a property of real numbers called zero-product rule.

Zero-Product Rule

If a and b are two numbers or expressions and if ab = 0, then either a = 0 or b = 0 or both a = 0 and b = 0

Using this rule, the solutions of the equation (x - 1)(x - 2) = 0 can be obtained by putting each factor equal to zero and then solving for x. Thus, we get

x - 1 = 0 or x - 2 = 0

⟹ x = 1 or x = 2.

Hence, the solutions of the equation \(x^2 - 3x + 2 = 0\) are 1 and 2.

Method To Solve A Quadratic Equation By Factorisation

Proceed As Under:

- Clear all fractions and write the equation in the form \(ax^2 + bx + c = 0\).

- Factorise the left hand side into product of two linear factors.

- Put each linear factor equal to zero and solve the resulting linear equations.

Remark

The solutions (roots) may be checked by substituting in the original equation.

Example 1.

Solve: \(x^2 - 2x - 15 = 0\).

Solution.

Given \(x^2 - 2x - 15 = 0\)

⟹ \(x^2 - 5x + 3x - 15 = 0\)

⟹ \(x(x - 5) + 3(x - 5) = 0\)

⟹ (x - 5)(x + 3) = 0

⟹ x - 5 = 0 or x + 3 = 0

⟹ x = 5 or x = -3.

Hence, the roots of the given equation are 5, -3.

Example 2.

Solve: \(6x^2 + x - 1 = 0\).

Solution.

Given \(6x^2 + x - 1 = 0\)

⟹ \(6x^2 + 3x - 2x - 1 = 0\)

⟹ \(3x(2x + 1) - 1(2x + 1) = 0\)

⟹ (2x + 1)(3x - 1) = 0

⟹ 2x + 1 = 0 or 3x - 1 = 0

⟹ \(x = -\frac{1}{2}\) or \(x = \frac{1}{3}\).

Hence, the roots of the given equation are \(-\frac{1}{2}, \frac{1}{3}\).

Example 3.

Solve: \(9x^2 + 6x + 1 = 0\).

Solution.

Given \(9x^2 + 6x + 1 = 0\)

⟹ \(9x^2 + 3x + 3x + 1 = 0\)

⟹ \(3x(3x + 1) + 1(3x + 1) = 0\)

⟹ (3x + 1)(3x + 1) = 0

⟹ 3x + 1 = 0 or 3x + 1 = 0

⟹ \(x = -\frac{1}{3}\) or \(x = -\frac{1}{3}\).

Hence, the roots of the given equation are \(-\frac{1}{3}, -\frac{1}{3}\).

Note that the given quadratic equation has equal roots.

Example 4.

Solve: \(2x - \frac{1}{x} = 1\).

Solution.

To clear the fractions, multiply both sides of the given equation by x. We get

\(2x^2 - 1 = x\)

⟹ \(2x^2 - x - 1 = 0\)

⟹ \(2x^2 - 2x + x - 1 = 0\)

⟹ \(2x(x - 1) + 1(x - 1) = 0\)

⟹ (2x + 1)(x - 1) = 0

⟹ 2x + 1 = 0 or x - 1 = 0

⟹ \(x = -\frac{1}{2}\) or x = 1.

Hence, the roots of the given equation are \(-\frac{1}{2}, 1\).

Example 5.

Solve: \(\frac{x + 3}{x - 1} = \frac{2x + 1}{3x - 5}\)

Solution.

Given \(\frac{x + 3}{x - 1} = \frac{2x + 1}{3x - 5}\)

⟹ (x + 3)(3x - 5) = (x - 1)(2x + 1) (By cross-multiplication)

⟹ \(3x^2 - 5x + 9x - 15 = 2x^2 + x - 2x - 1\)

⟹ \(3x^2 + 4x - 15 = 2x^2 - x - 1\)

⟹ \(3x^2 - 2x^2 + 4x + x - 15 + 1 = 0\)

⟹ \(x^2 + 5x - 14 = 0\)

⟹ \(x^2 + 7x - 2x - 14 = 0\)

⟹ \(x(x + 7) - 2(x + 7) = 0\)

⟹ (x + 7)(x - 2) = 0

⟹ x + 7 = 0 or x - 2 = 0

⟹ x = -7 or x = 2.

Hence, the roots of the given equation are -7, 2.

Example 6.

Solve \(\frac{x}{x - 1} + \frac{x - 1}{x} = 2\frac{1}{2}\).

Solution.

Given \(\frac{x}{x - 1} + \frac{x - 1}{x} = \frac{5}{2}\)

To clear the fractions, multiply both sides of the given equation by L.C.M. of denominators i.e. by 2x(x - 1). We get

2x \times x + 2(x - 1)(x - 1) = 5x(x - 1)

⟹ \(2x^2 + 2x^2 - 4x + 2 = 5x^2 - 5x\)

⟹ \(4x^2 - 4x + 2 = 5x^2 - 5x\)

⟹ \(4x^2 - 5x^2 - 4x + 5x + 2 = 0\)

⟹ \(-x^2 + x + 2 = 0\)

⟹ \(x^2 - x - 2 = 0\)

⟹ \(x^2 - 2x + x - 2 = 0\)

⟹ \(x(x - 2) + 1(x - 2) = 0\)

⟹ (x + 1)(x - 2) = 0

⟹ x + 1 = 0 or x - 2 = 0

⟹ x = -1 or x = 2.

Hence, the roots of the given equation are -1, 2.

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