ICSE Class 8 Maths Chapter 20 Quadratic Equation

Chapter 20: Quadratic Equation

20.1 Introduction

An equation, which contains one unknown (variable) with highest power 2, is called a quadratic equation.

In other words, a quadratic equation is a second degree equation in one variable.

e.g. (i) \(x^2 - 5x + 7 = 0\) (ii) \(5x^2 - 7x = 0\) (iii) \(3x^2 - 8 = 0\) and so on.

The standard form of a quadratic equation is \(ax^2 + bx + c = 0\); where a, b and c are all real numbers and \(a \neq 0\).

Remember

1. In quadratic equation \(ax^2 + bx + c = 0\); the term \(ax^2\) contains the second power of the unknown and the term bx contains the first power of the unknown. The term c which does not contain x is called the absolute term.

2. A quadratic equation, which contains both the first and the second powers of the unknown quantity, is called a complete quadratic equation.

e.g. (i) \(x^2 + 5x - 6 = 0\) (ii) \(3x^2 - 7x + 8 = 0\) and so on.

3. A quadratic equation, which contains only the second power of the unknown quantity, is called an incomplete quadratic equation.

e.g. (i) \(x^2 - 16 = 0\) (ii) \(5x^2 = 20\) (iii) \(7x^2 - 8 = 0\) and so on.

4. A quadratic equation gives two values of the unknown (variable) and both these values are called the roots of the equation.

Test Yourself

1. \(5x + 7 = 0\) is a linear equation containing only one variable.

2. \(5x + 7y = 8\) is a linear equation containing two variables.

3. \(4x^2 - 7x + 3 = 0\) is a quadratic equation containing only one variable.

4. The degree of equation \(4x^2 - 7x + 3 = 0\) is 2; and it always gives 2 values of its variable x.

5. \(4x^2 - 1 = 0\) is a quadratic equation and so it will give 2 values of its variable.

6. \(3x^2 - 5x = 0\) is a quadratic equation.

7. \((3x - 5)(4x + 3) = 0 \Rightarrow\) __________ = 0, or __________ = 0 \(\Rightarrow x =\) __________ or \(x =\) __________

20.2 Solving A Quadratic Equation

Steps

1. Express the equation in the form \(ax^2 + bx + c = 0\).

2. Factorise the expression on the left hand side of the resulting equation.

3. Put each of the factors equal to zero and solve.

Example 1

Solve: \(2x^2 - 3 = 5x\)

Solution

Step 1: Converting into standard form:

\(2x^2 - 3 = 5x\)

\(\Rightarrow 2x^2 - 5x - 3 = 0\) [Standard form]

Step 2: Factorising left hand side:

\(2x^2 - 6x + x - 3 = 0 \Rightarrow 2x(x - 3) + 1(x - 3) = 0\)

\(\Rightarrow (x - 3)(2x + 1) = 0\)

Step 3: Putting each of the factors equal to zero:

i.e. \(x - 3 = 0\) or \(2x + 1 = 0\)

\(\Rightarrow x = 3\) or \(2x = -1\) [Solving equations]

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

Example 2

(i) \(x^2 = 25\) (ii) \(x + \frac{1}{x} = 2\frac{1}{2}\)

Solution

(i) \(x^2 - 25 = 0 \Rightarrow (x - 5)(x + 5) = 0\) [Step 2]

\(\Rightarrow x - 5 = 0\) or \(x + 5 = 0\) [Step 3]

\(\Rightarrow x = 5\) or \(x = -5\) (Ans.)

(ii) \(x + \frac{1}{x} = 2\frac{1}{2} \Rightarrow \frac{x^2 + 1}{x} = \frac{5}{2}\)

\(\Rightarrow 2x^2 + 2 = 5x\)

\(\Rightarrow 2x^2 - 5x + 2 = 0\) [Step 1]

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

\(\Rightarrow 2x(x - 2) - 1(x - 2) = 0\)

\(\Rightarrow (x - 2)(2x - 1) = 0\) [Step 2]

\(\Rightarrow x - 2 = 0\) or \(2x - 1 = 0\) [Step 3]

\(\Rightarrow x = 2\) or \(x = \frac{1}{2}\) (Ans.)

Example 3

(i) \(x^2 - 5x = 0\) (ii) \(\frac{3x - 7}{2x - 5} = \frac{x + 1}{x - 1}\)

Solution

(i) \(x^2 - 5x = 0 \Rightarrow x(x - 5) = 0\) [Step 2]

\(\Rightarrow x = 0\) or \(x - 5 = 0\) [Step 3]

\(\Rightarrow x = 0\) or \(x = 5\) (Ans.)

(ii) \(\frac{3x - 7}{2x - 5} = \frac{x + 1}{x - 1} \Rightarrow (3x - 7)(x - 1) = (2x - 5)(x + 1)\) [By cross-multiplying]

\(\Rightarrow 3x^2 - 3x - 7x + 7 = 2x^2 + 2x - 5x - 5\)

\(\Rightarrow 3x^2 - 10x + 7 - 2x^2 + 3x + 5 = 0\)

\(\Rightarrow x^2 - 7x + 12 = 0\) [Step 1]

\(\Rightarrow x^2 - 4x - 3x + 12 = 0\)

\(\Rightarrow x(x - 4) - 3(x - 4) = 0\)

\(\Rightarrow (x - 4)(x - 3) = 0\) [Step 2]

\(\Rightarrow x - 4 = 0\) or \(x - 3 = 0\) [Step 3]

\(\Rightarrow x = 4\) or \(x = 3\) (Ans.)

Teacher's Note

Quadratic equations model many real-world situations like calculating areas, predicting profits, or determining the trajectory of a ball - understanding these equations helps us solve practical problems in business, engineering, and sports.

Example 4

Solve: \(\frac{1}{x} + \frac{1}{2 + x} = \frac{3}{4}\)

Solution

\(\frac{1}{x} + \frac{1}{2 + x} = \frac{3}{4} \Rightarrow \frac{2 + x + x}{x(2 + x)} = \frac{3}{4}\)

\(\Rightarrow \frac{2 + 2x}{2x + x^2} = \frac{3}{4}\)

\(\Rightarrow 3(2x + x^2) = 4(2 + 2x)\)

\(\Rightarrow 6x + 3x^2 = 8 + 8x\) i.e. \(3x^2 - 2x - 8 = 0\)

\(\Rightarrow 3x^2 - 6x + 4x - 8 = 0\) i.e. \(3x(x - 2) + 4(x - 2) = 0\)

\(\Rightarrow (x - 2)(3x + 4) = 0\) i.e. \(x - 2 = 0\) or \(3x + 4 = 0\)

\(\Rightarrow x = 2\) or \(x = -\frac{4}{3}\) (Ans.)

Test Yourself

8. \(x(x - 3) = 0 \Rightarrow\) ________ = 0 or ________ = 0 \(\Rightarrow x =\) ________ or \(x =\) ________

9. \(4x^2 - 25 = 0 \Rightarrow\) __________ = 0 \(\Rightarrow\) __________ = 0 or __________ = 0 \(\Rightarrow x =\) ...... or \(x =\) ......

10. \(x = \frac{9}{x} \Rightarrow x^2 =\) __________ and \(x =\) __________

11. \((x - 3)^2 = 36 \Rightarrow x - 3 =\) __________\(\Rightarrow x - 3 =\) __________, or \(x - 3 =\) __________

\(\Rightarrow x =\) __________ or \(x =\) __________

12. \(x(x - 5) = 6 \Rightarrow\) ______________________________________________________________________________

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