ICSE Class 8 Maths Algebra Chapter 22 Quadratic Equations PDF Download

Quadratic Equations

Quadratic Equation

Solving and Framing Quadratic Equation

Introduction

A simple linear equation is an algebraic statement of equality that involves only one variable, the degree of which is not more than 1. Solving a simple linear equation gives us the one and only value of the variable that satisfies the equation.

A quadratic equation is an algebraic statement of equality that involves only one variable, the degree of which is not more than 2. Solving a quadratic equation gives us two values of the variable, both of which satisfy the equation.

Quadratic equations are of two forms.

1. In the pure form of quadratic equations, the degree of the variable is always 2. Examples of pure quadratic equations are: \(x^2 = 25, 3x^2 = 392\)

2. In the standard form of quadratic equations, the degree of the variable is 2 as well as 1. Examples of the standard form of quadratic equations are: \(x^2 - 4x - 32 = 0, 6x^2 + x - 12 = 0\)

Solving Quadratic Equations

Example 1: Solve \(6x^2 = 384\)

\(\Rightarrow \quad x^2 = \frac{384}{6} = 64\)

\(\Rightarrow \quad x^2 - 64 = 0\)

\(\Rightarrow \quad (x + 8)(x - 8) = 0\)

Thus, either \(x + 8 = 0\), or \(x - 8 = 0\)

when \(x + 8 = 0 \Rightarrow x = -8\)

when \(x - 8 = 0 \Rightarrow x = 8\)

Thus, the roots of \(6x^2 = 384\) are \(\pm 8\).

Example 2: Solve \(16y^2 = 25\)

\(\Rightarrow \quad y^2 = \frac{25}{16}\)

\(\Rightarrow \quad y^2 - \frac{25}{16} = 0\)

\(\Rightarrow \quad \left(y + \frac{5}{4}\right)\left(y - \frac{5}{4}\right) = 0\)

Thus, either \(y + \frac{5}{4} = 0\), or \(y - \frac{5}{4} = 0\)

when \(y + \frac{5}{4} = 0 \Rightarrow y = -\frac{5}{4} = -1\frac{1}{4}\)

when \(y - \frac{5}{4} = 0 \Rightarrow y = \frac{5}{4} = 1\frac{1}{4}\)

Thus, the roots of \(16y^2 = 25\) are \(\pm 1\frac{1}{4}\).

Example 3: Solve \(2x^2 + 9x = 35\)

Step 1: Move all terms to LHS leaving 0 on the RHS to convert the given equation to the standard form.

\(\Rightarrow \quad 2x^2 + 9x - 35 = 0\)

Step 2: Factorise the quadratic trinomial on LHS.

\(\Rightarrow \quad 2x^2 + 14x - 5x - 35 = 0\)

\(\Rightarrow \quad 2x(x + 7) - 5(x + 7) = 0\)

\(\Rightarrow \quad (2x - 5)(x + 7) = 0\)

Step 3: Treat each factor as a simple linear equation to solve and find the roots of the quadratic equation.

\(2x - 5 = 0\) or \(x + 7 = 0\)

\(\Rightarrow \quad 2x = 5\) or \(\Rightarrow \quad x = -7\)

\(\Rightarrow \quad x = \frac{5}{2}\) or \(x = -7\)

Thus, the roots of \(2x^2 + 9x = 35\) are \(\frac{5}{2}\) and \(-7\).

CHECK: When \(x = \frac{5}{2} = 2.5\)

\((2 \times 2.5 \times 2.5) + (9 \times 2.5) = 35\)

\(\Rightarrow \quad 12.5 + 22.5 = 35\)

\(\Rightarrow \quad 35 = 35\)

When \(x = -7\)

\((2 \times -7 \times -7) + (9 \times -7) = 35\)

\(\Rightarrow \quad 98 - 63 = 35\)

\(\Rightarrow \quad 35 = 35\)

The roots obtained satisfy the given quadratic equation.

Example 4: Solve \(x^2 - x - 30 = 0\)

\(\Rightarrow \quad x^2 - 6x + 5x - 30 = 0\)

\(\Rightarrow \quad x(x - 6) + 5(x - 6) = 0\)

\(\Rightarrow \quad (x + 5)(x - 6) = 0\)

\(\Rightarrow \quad x + 5 = 0\) or \(x - 6 = 0\)

\(\Rightarrow \quad x = -5\) or \(x = 6\)

Thus, the roots of \(x^2 - x - 30 = 0\) are \(-5\) and \(6\).

Example 5: Solve \(\frac{7x}{x^2 + 10} = \frac{6}{x + 1}\)

\(\Rightarrow \quad 7x(x + 1) = 6(x^2 + 10)\)

\(\Rightarrow \quad 7x^2 + 7x = 6x^2 + 60\)

\(\Rightarrow \quad 7x^2 - 6x^2 + 7x - 60 = 0\)

\(\Rightarrow \quad x^2 + 7x - 60 = 0\)

\(\Rightarrow \quad x^2 + 12x - 5x - 60 = 0\)

\(\Rightarrow \quad x(x + 12) - 5(x + 12) = 0\)

\(\Rightarrow \quad (x - 5)(x + 12) = 0\)

\(\Rightarrow \quad x - 5 = 0\) or \(x + 12 = 0\)

\(\Rightarrow \quad x = 5\) or \(x = -12\)

Thus, the roots of \(\frac{7x}{x^2 + 10} = \frac{6}{x + 1}\) are \(5\) and \(-12\).

Teacher's Note

Quadratic equations model real-world situations like calculating projectile motion or finding the optimal dimensions for a garden, making algebra a practical tool for everyday problem-solving.

Example 6: Solve \(\frac{x + 3}{3x - 5} = \frac{2x - 5}{x - 1}\)

\(\Rightarrow \quad (x + 3)(x - 1) = (2x - 5)(3x - 5)\)

\(\Rightarrow \quad x^2 - x + 3x - 3 = 6x^2 - 10x - 15x + 25\)

\(\Rightarrow \quad x^2 + 2x - 3 = 6x^2 - 25x + 25\)

\(\Rightarrow \quad 6x^2 - 25x + 25 - x^2 - 2x + 3 = 0\)

\(\Rightarrow \quad 5x^2 - 27x + 28 = 0\)

\(\Rightarrow \quad 5x^2 - 20x - 7x + 28 = 0\)

\(\Rightarrow \quad 5x(x - 4) - 7(x - 4) = 0\)

\(\Rightarrow \quad (5x - 7)(x - 4) = 0\)

\(\Rightarrow \quad 5x - 7 = 0\) or \(x - 4 = 0\)

\(\Rightarrow \quad 5x = 7\) or \(x = 4\)

\(\Rightarrow \quad x = \frac{7}{5}\) or \(x = 4\)

Thus, the roots of \(\frac{x + 3}{3x - 5} = \frac{2x - 5}{x - 1}\) are \(4\) and \(\frac{7}{5}\).

Some equations can be reduced to the standard form of quadratic equations by simple substitution.

Example 7: Solve \((a - 2)^2 - 6 = a - 2\)

Let \(a - 2 = x\), then we have

\(x^2 - 6 = x\)

\(\Rightarrow \quad x^2 - x - 6 = 0\)

\(\Rightarrow \quad x^2 - 3x + 2x - 6 = 0\)

\(\Rightarrow \quad x(x - 3) + 2(x - 3) = 0\)

\(\Rightarrow \quad (x + 2)(x - 3) = 0\)

\(\Rightarrow \quad x + 2 = 0\) or \(x - 3 = 0\)

\(\Rightarrow \quad x = -2\) or \(x = 3\)

As \(a - 2 = x\),

\(\Rightarrow \quad a - 2 = -2\) or \(a - 2 = 3\)

\(\Rightarrow \quad a = 0\) or \(a = 5\)

Example 8: Solve \(a^4 - 25a^2 + 144 = 0\)

Let \(a^2 = x\), then we have

\(x^2 - 25x + 144 = 0\)

\(\Rightarrow \quad x^2 - 9x - 16x + 144 = 0\)

\(\Rightarrow \quad x(x - 9) - 16(x - 9) = 0\)

\(\Rightarrow \quad (x - 16)(x - 9) = 0\)

\(\Rightarrow \quad x - 16 = 0\) or \(x - 9 = 0\)

\(\Rightarrow \quad x = 16\) or \(x = 9\)

As \(a^2 = x\),

\(\Rightarrow \quad a^2 = 16\) or \(a^2 = 9\)

\(\Rightarrow \quad a = \pm 4\) or \(a = \pm 3\)

Teacher's Note

Using substitution to simplify complex equations teaches students how breaking down difficult problems into smaller, manageable parts helps solve real challenges efficiently.

Framing a Quadratic Equation

A quadratic equation is solved by expressing it as a product of factors being equal to 0. As a reverse process, given the roots, we can frame a quadratic equation as shown in the following examples.

Example 9: Frame the quadratic equation whose roots are 2 and -4.

Given \(x = 2\) or \(x = -4\)

\(\Rightarrow \quad x - 2 = 0\) or \(x + 4 = 0\)

Multiplying the two equations, we have

\((x - 2)(x + 4) = 0\)

\(\Rightarrow \quad x^2 + 4x - 2x - 8 = 0\)

\(\Rightarrow \quad x^2 + 2x - 8 = 0\)

Example 10: Frame the quadratic equation whose roots are \(-\frac{3}{5}\) and \(\frac{2}{7}\).

Given \(x = -\frac{3}{5}\) or \(x = \frac{2}{7}\)

\(\Rightarrow \quad 5x = -3\) or \(7x = 2\)

\(\Rightarrow \quad 5x + 3 = 0\) or \(7x - 2 = 0\)

Multiplying the two equations, we have

\((5x + 3)(7x - 2) = 0\)

\(\Rightarrow \quad 35x^2 - 10x + 21x - 6 = 0\)

\(\Rightarrow \quad 35x^2 + 11x - 6 = 0\)

Teacher's Note

Framing equations from roots is like reverse-engineering: if we know the answer, can we build the question? This critical thinking skill applies to many professional fields and scientific research.

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