ICSE Class 10 Maths Chapter 05 Quadratic Equations

Quadratic Equations

Introduction

An equation with one variable, in which the highest power of the variable is two, is known as a quadratic equation.

For example:

(i) \(3x^2 + 4x + 7 = 0\)

(ii) \(4x^2 + 5x = 0\)

(iii) \(2x^2 - 50 = 0\)

(iv) \(x^2 = 4\), etc.

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

e.g. equation \(4x^2 + 5x - 6 = 0\) is a quadratic equation in standard form.

2. Every quadratic equation gives two values of the unknown variable used in it and these values are called roots of the equation.

3. Discriminant: For the quadratic equation \(ax^2 + bx + c = 0\), \(a \ne 0\); the expression \(b^2 - 4ac\) is called discriminant and is, in general, denoted by the letter D.

Thus, discriminant \(D = b^2 - 4ac\).

4. If a quadratic equation contains only two terms one square term and one first power term of the unknown, it is called affected quadratic equation.

For example: (i) \(4x^2 + 5x = 0\) (ii) \(7x^2 - 3x = 0\), etc.

5. If the quadratic equation contains only the square of the unknown, it is called pure quadratic equation.

For example: (i) \(x^2 = 4\) (ii) \(3x^2 - 8 = 0\), etc.

To Examine the Nature of the Roots

Examining the roots of a quadratic equation means to know the type of its roots i.e. whether they are real or imaginary, rational or irrational, equal or unequal.

The nature of the roots of a quadratic equation depends entirely on the value of its discriminant \(b^2 - 4ac\).

If for a quadratic equation \(ax^2 + bx + c = 0\); where a, b and c are real numbers and \(a \ne 0\), then discriminant:

(i) \(b^2 - 4ac = 0\) \(\Rightarrow\) the roots are real and equal.

(ii) \(b^2 - 4ac > 0\) \(\Rightarrow\) the roots are real and unequal.

(iii) \(b^2 - 4ac < 0\) \(\Rightarrow\) the roots are imaginary (not real).

Teacher's Note

Understanding how to determine if a quadratic equation has real or imaginary roots helps in practical situations like calculating projectile motion in sports or determining if a business investment will have real returns.

1. Every number, whether it is rational or irrational, is a real number. i.e.

(i) every rational number is a real number and

(ii) every irrational number is also a real number.

2. Square root of a negative number is an imaginary number.

Thus: each of \(\sqrt{-4}\), \(\sqrt{-8}\), \(2\sqrt{-5}\), ......, etc. is an imaginary number.

Without solving, examine the nature of the roots of the equations:

(i) \(5x^2 - 6x + 7 = 0\) (ii) \(x^2 + 6x + 9 = 0\) (iii) \(2x^2 + 6x + 3 = 0\)

Solution

(i) Comparing given quadratic equation \(5x^2 - 6x + 7 = 0\) with equation \(ax^2 + bx + c = 0\); we get: \(a = 5\), \(b = -6\) and \(c = 7\).

\(\Rightarrow\) Discriminant \(= b^2 - 4ac = (-6)^2 - 4 \times 5 \times 7\)

\(= 36 - 140 = -104\); which is negative.

Since, a, b and c are real numbers; \(a \ne 0\) and \(b^2 - 4ac < 0\).

\(\therefore\) The roots are not real i.e. the roots are imaginary. Ans.

(ii) Comparing quadratic equation \(x^2 + 6x + 9 = 0\) with \(ax^2 + bx + c = 0\); we get: \(a = 1\), \(b = 6\) and \(c = 9\)

\(\Rightarrow\) \(b^2 - 4ac = (6)^2 - 4 \times 1 \times 9 = 36 - 36 = 0\)

Since, a, b and c are real numbers; \(a \ne 0\) and \(b^2 - 4ac = 0\).

\(\therefore\) The roots are real and equal. Ans.

(iii) Comparing \(2x^2 + 6x + 3 = 0\) and \(ax^2 + bx + c\), we get: \(a = 2\), \(b = 6\) and \(c = 3\)

\(b^2 - 4ac = (6)^2 - 4 \times 2 \times 3\)

\(= 36 - 24 = 12\); which is positive.

Since, a, b and c are real numbers; \(a \ne 0\) and \(b^2 - 4ac > 0\).

\(\therefore\) The roots are real and unequal. Ans.

Teacher's Note

The discriminant test is similar to checking a recipe's ingredients before cooking - you can predict the outcome before actually doing the work, saving time and resources.

Find the value of m, if the roots of the following quadratic equation are equal: \((4 + m)x^2 + (m + 1)x + 1 = 0\).

Solution

For the given equation \((4 + m)x^2 + (m + 1)x + 1 = 0\);

\(a = 4 + m\), \(b = m + 1\) and \(c = 1\)

Since, the roots are equal

\(\therefore\) \(b^2 - 4ac = 0\) \(\Rightarrow\) \((m + 1)^2 - 4(4 + m) \times 1 = 0\)

\(\Rightarrow\) \(m^2 + 2m + 1 - 16 - 4m = 0\)

\(\Rightarrow\) \(m^2 - 2m - 15 = 0\)

On solving, we get: \(m = 5\) or \(m = -3\) Ans.

Exercise 5(A)

1. Without solving, comment upon the nature of roots of each of the following equations:

(i) \(7x^2 - 9x + 2 = 0\) (ii) \(6x^2 - 13x + 4 = 0\)

(iii) \(25x^2 - 10x + 1 = 0\) (iv) \(x^2 + 2\sqrt{3}x - 9 = 0\)

(v) \(x^2 - ax - b^2 = 0\) (vi) \(2x^2 + 8x + 9 = 0\)

2. Find the value of p, if the following quadratic equations have equal roots:

(i) \(4x^2 - (p - 2)x + 1 = 0\)

(ii) \(x^2 + (p - 3)x + p = 0\) [2013]

3. The equation \(3x^2 - 12x + (n - 5) = 0\) has equal roots. Find the value of n.

4. Find the value of m, if the following equation has equal roots: \((m - 2)x^2 - (5 + m)x + 16 = 0\)

5. Find the value of k for which the equation \(3x^2 - 6x + k = 0\) has distinct and real root. [2015]

Solving Quadratic Equations by Factorisation

Steps: (i) Clear all fractions and brackets, if necessary.

(ii) Transpose all the terms to the left hand side to get an equation in the form \(ax^2 + bx + c = 0\).

(iii) Factorise the expression on the left hand side.

(iv) Put each factor equal to zero and solve.

Zero Product Rule: Whenever the product of two expressions is zero; at least one of the expressions is zero.

Thus, if \((x + 3) (x - 2) = 0\)

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

\(\Rightarrow\) \(x = -3\), or \(x = 2\).

Solve: (i) \(2x^2 - 7x = 39\) (ii) \(x^2 = 5x\) (iii) \(x^2 = 16\)

Solution

(i) \(2x^2 - 7x = 39\)

\(\Rightarrow\) \(2x^2 - 7x - 39 = 0\) [Expressing as \(ax^2 + bx + c = 0\)]

\(\Rightarrow\) \(2x^2 - 13x + 6x - 39 = 0\) [Factorising the left hand side]

\(\Rightarrow\) \(x (2x - 13) + 3 (2x - 13) = 0\)

\((2x - 13) (x + 3) = 0\)

\(2x - 13 = 0\), or \(x + 3 = 0\) [Zero Product Rule]

\(\Rightarrow x = \frac{13}{2}\), or \(x = -3\) Ans.

(ii) \(x^2 = 5x\) \(\Rightarrow\) \(x^2 - 5x = 0\)

\(\Rightarrow\) \(x (x - 5) = 0\)

\(\Rightarrow\) \(x = 0\), or \(x - 5 = 0\)

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

(iii) \(x^2 = 16\) \(\Rightarrow\) \(x^2 - 16 = 0\)

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

\(\Rightarrow\) \(x + 4 = 0\), or \(x - 4 = 0\)

\(\Rightarrow\) \(x = - 4\), or \(x = 4\) Ans.

Alternative method:

\(x^2 = 16\)

\(\Rightarrow\) \(x = \pm 4\)

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

Teacher's Note

Factorisation mirrors how we break down complex problems in real life - by finding the simplest components that when combined reproduce the original problem, we can solve it step by step.

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