Compulsory Mathematics · Chapter 058 periods4 sections · 6 questions

Quadratic Equations

द्विघात समीकरण

Introduction and Standard Form3 Solving by Factorization3 The Quadratic Formula4 Nature of Roots (Discriminant)2

Introduction and Standard Form

An equation of the form $a x^{2} + b x + c = 0$ , where $a, b, c$ are real numbers and $a \neq = 0$ , is called a quadratic equation. The highest power of the variable $x$ is 2.

Standard form

a x^{2} + b x + c = 0, a \neq = 0

For example, $x^{2} - 5 x + 6 = 0$ and $2 x^{2} - 3 x - 5 = 0$ are quadratic equations, but $3 x + 7 = 0$ is not, since the highest power of $x$ is 1.

Solving by Factorization

If a quadratic expression can be written as a product of two linear factors, we can find the roots by setting each factor equal to zero. We split the middle term $b x$ into two terms whose coefficients multiply to $a c$ and add to $b$ .

Worked Example

Solve $x^{2} - 5 x + 6 = 0$ by factorization.

1We need two numbers whose product is $6$ and sum is $- 5$ : these are $- 2$ and $- 3$ .
2 $x^{2} - 2 x - 3 x + 6 = 0 \Rightarrow x (x - 2) - 3 (x - 2) = 0 \Rightarrow (x - 2) (x - 3) = 0$
3So $x - 2 = 0$ or $x - 3 = 0$ , giving $x = 2$ or $x = 3$ .

Answer

x = 2, 3

Upward parabola crossing x-axis at x=2 and x=3, vertex below x-axis between the roots — Graph of $y = x^{2} - 5 x + 6$ showing the parabola crossing the x-axis at the roots

The Quadratic Formula

When factorization is difficult, the roots of $a x^{2} + b x + c = 0$ can always be found using the quadratic formula, derived by completing the square.

Quadratic formula

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

A square of side x with an attached rectangle of width b split into two strips, rearranged into a larger square with a small missing corner — Geometric idea of completing the square for $x^{2} + b x$

Worked Example

Solve $2 x^{2} - 3 x - 5 = 0$ using the quadratic formula.

1Here $a = 2$ , $b = - 3$ , $c = - 5$ .
2 $x = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 ( 2 ) ( - 5 )}{2 ( 2 )} = \frac{3 \pm 9 + 40}{4} = \frac{3 \pm 49}{4} = \frac{3 \pm 7}{4}$
3 $x = \frac{10}{4} = \frac{5}{2}$ or $x = \frac{- 4}{4} = - 1$

Answer

x = \frac{5}{2}, - 1

Nature of Roots (Discriminant)

The expression $D = b^{2} - 4 a c$ is called the discriminant. It tells us the nature of the roots without solving the equation fully.

If $D > 0$ : two distinct real roots.
If $D = 0$ : two equal real roots.
If $D < 0$ : no real roots (roots are imaginary).

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Practice Questions

Press “Show solution” on any question to reveal its step-by-step bilingual answer.

13 marks

Solve by factorization: $x^{2} - 7 x + 12 = 0$

24 marks

Using the quadratic formula, solve $3 x^{2} + 5 x - 2 = 0$

34 marksMulti-part

Find the discriminant of $2 x^{2} - 4 x + 5 = 0$ 2 marks

Hence state the nature of its roots.2 marks

45 marksMulti-part

The product of two consecutive positive integers is 132. Form a quadratic equation to represent this statement, taking the smaller integer as $x$ .2 marks

Solve the equation formed in part (a) to find the two integers.3 marks

54 marks

Solve for $x$ : $\frac{x}{x - 1} + \frac{x - 1}{x} = 2 \frac{1}{2}$

63 marks

From the graph shown above (Figure 5.1), write down the roots of $x^{2} - 5 x + 6 = 0$ and verify them by substitution.

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