Nature of Roots of a Quadratic Equation

First of all, let’s recall the quadratic formula. Yes, in fact, the nature of roots can be determined by simply the value of an expression. Do you remember what nature of roots of a quadratic equation is? Does it refer to whether the roots are real or not real, equal or unequal?

Discriminant of a Quadratic Equation Consider a quadratic equation: ax2 + bx + c = 0, where a  0 The value of this expression can determine the nature of roots. a 2 ac 4 b x - ± = Its roots are given by:  Quadratic formula The expression is called the discriminant (denoted by ).  pronounced as ‘delta’ Discriminant:  = b2  4ac

ac b 4 - Case 1:  > 0  i.e. b2 – 4ac > 0 ∵ is a positive real number. ∵ ∴ The roots of the quadratic equation are and . a ac b 2 4 - +  The two roots are real and unequal. two unequal real roots

ac b 4 - Case 1:  > 0  i.e. b2 – 4ac > 0 ∵ is a positive real number. ∵ ∴ The roots of the quadratic equation are and . a ac b 2 4 - +  ∴ The equation has two unequal real roots. e.g. Consider x2 + 3x – 7 = 0. = 32 – 4(1)(–7) = 37 > 0 ∴ The equation x2 + 3x – 7 = 0 has two unequal real roots.

ac b 4 - Case 2:  = 0  i.e. b2 – 4ac = 0 ∵ is zero. ∴ The roots of the quadratic equation are . a b 2 - . a b 2  - ac b 4 2 - The two roots are real and equal. one double real root

ac b 4 - Case 2:  = 0  i.e. b2 – 4ac = 0 ∵ is zero. ∴ The roots of the quadratic equation are . a b 2 - ∴ The equation has one double real root. e.g. Consider x2 – 8x + 16 = 0. = (–8)2 – 4(1)(16) = 0 ∴ The equation x2 – 8x + 16 = 0 has one double real root.

ac b 4 - Case 3:  < 0  i.e. b2 – 4ac < 0 is not a real number. ∵ ∴ The roots of the quadratic equation are not real. ∴ The equation has no real roots. The two roots are no real roots.

ac b 4 - Case 3:  < 0  i.e. b2 – 4ac < 0 is not a real number. ∵ ∴ The roots of the quadratic equation are not real. ∴ The equation has no real roots. e.g. Consider x2 – 2x + 5 = 0. = (–2)2 – 4(1)(5) = –16 < 0 ∴ The equation x2 – 2x + 5 = 0 has no real roots.

The table below summarizes the three cases of the nature of roots of a quadratic equation. Condition Nature of its roots  > 0  = 0  < 0 two unequal real roots one double real root no real roots (or two distinct real roots) (or two equal real roots)

Follow-up question Find the value of the discriminant of the equation x2 – 5x + 3 = 0, and hence determine the nature of its roots. The discriminant of x2 – 5x + 3 = 0 is given by: 13 ) 3 )( 1 ( 4 5 2 = - D > ∴ The equation has two distinct real roots.

Graph of a Quadratic Equation For the quadratic equation ax2 + bx + c = 0, are there any relations among the following: discriminant, nature of roots, no. of x-intercepts of the corresponding graph.

For the quadratic equation ax2 + bx + c = 0: 2 unequal real roots 1 double real root 0 real roots Value of the discriminant determine nature of roots of ax2 + bx + c = 0 no. of x-intercepts of the graph of y = ax2 + bx + c 2 x-intercepts 1 x-intercept 0 x-intercepts Roots of ax2 + bx + c = 0 equals no. of x-intercepts of the graph of y = ax2 + bx + c value of the discriminant Therefore also tells us

No. of x-intercepts of the graph of y = ax 2 + bx + c The discriminant of the quadratic equation ax2 + bx + c = 0, the nature of its roots and the number of x-intercepts of the graph of y = ax2 + bx + c have the following relations. Discriminant ( = b 2  4ac) Nature of roots of ax 2 + bx + c = 0 No. of x-intercepts of the graph of y = ax 2 + bx + c 2 unequal real roots 1 double real root no real roots 2 1  > 0  = 0  < 0

Follow-up question Some graphs of y = ax2 + bx + c will be shown one by one. For each corresponding quadratic equation ax2 + bx + c = 0, determine whether  > 0,  = 0 or  < 0. O y x O y x  > 0 2 x-intercepts  > 0 2 x-intercepts O y x O x y 1 x-intercept  < 0  = 0 no x-intercepts

Forming a Quadratic Equation with Given Roots

Miss Chan, I have learnt how to solve a quadratic equation, but how can I form a quadratic equation from two given roots? We can form the equation by reversing the process of solving a quadratic equation by the factor method.

Consider the following example: Solve an equation x2 – 4x + 3 = 0 (x – 1)(x – 3) = 0 x – 1 = 0 or x – 3 = 0 x = 1 or x = 3

Form an equation from two given roots Consider the following example: Form an equation from two given roots (x – 1)(x – 3) = 0 Reversing the process x – 1 = 0 or x – 3 = 0 x = 1 or x = 3

∴ The quadratic equation is x2 – 4x + 3 = 0. Now, let’s study the steps of forming a quadratic equation from given roots (1 and 3) again. x = 1 or x = 3 Note this key step. x – 1 = 0 or x – 3 = 0 (x – 1)(x – 3) = 0 x2 – 4x + 3 = 0 ∴ The quadratic equation is x2 – 4x + 3 = 0.

In general, Roots , (x – )(x – ) = 0 Quadratic Equation α β α β

For example, Roots , (x – )(x – ) = 0 Quadratic Equation 1 2 1 2

, (x – )(x – ) = 0 –4 (x – )[x – ] = 0 α 1 (–4) β 2 For example, Roots Quadratic Equation –4 Quadratic Equation (x – )[x – ] = 0 α 1 (–4) β 2

In general, If α and β are the roots of a quadratic equation in x, then the equation is: (x – α)(x – β) = 0

Follow-up question In each of the following, form a quadratic equation in x with the given roots, and write the equation in the general form. (a) –1, –5 (b) (a) The required quadratic equation is [x – (–1)][x – (–5)] = 0 (x + 1)(x + 5) = 0 x2 + x + 5x + 5 = 0 x2 + 6x + 5 = 0

(b) The required quadratic equation is

but it is too tedious to expand the left hand side of the equation I am trying to form a quadratic equation whose roots are and , In fact, there is another method to form a quadratic equation. It helps you form this quadratic equation.

Using Sum and Product of Roots Suppose α and β are the roots of a quadratic equation. Then, the equation can be written as: (x – )(x – ) = 0 x2 – x – x +  = 0 By expansion x2 – ( + )x +  = 0 Sum of roots Product of roots

Using Sum and Product of Roots Suppose α and β are the roots of a quadratic equation. Then, the equation can be written as: x2 – (sum of roots)x + product of roots = 0

Let’s find the quadratic equation whose roots are and . ∴ The required quadratic equation is x2 – (sum of roots)x + product of roots = 0

Follow-up question Form a quadratic equation in x whose roots are and . (Write your answer in the general form.)

Follow-up question Form a quadratic equation in x whose roots are and . (Write your answer in the general form.) ∴ The required quadratic equation is

Sum and Product of Roots

Relations between Roots and Coefficients Suppose  and  are the roots of ax2 + bx + c = 0. We can express  +  and  in terms of a, b and c.

Compare the coefficient of x and the constant term. ax2 + bx + c = 0 (x – )(x – ) = 0 2 = + a c x b x2 – x – x +  = 0 x2 – ( + )x +  = 0 Compare the coefficient of x and the constant term. Sum of roots = Product of roots = – a b c  =  +  =

For each of the following quadratic equations, find the sum and the product of its roots. Sum of roots = Product of roots = 2 7 – = 2x2 + 7x = 0 2x2 + 7x + 0 = 0 3x – x2 = 1 Sum of roots = Product of roots = 3 1 ) ( = – x2 – 3x + 1 = 0

Follow-up question It is given that the sum of the roots of x2 – (3 – 4k)x – 6k = 0 is –9. (a) Find the value of k. (b) Find the product of the roots. For the equation x2 – (3 – 4k)x – 6k = 0, (a) sum of roots = a b – Sum of roots = = 3 – 4k ∴ –9 = 3 – 4k k = 3

Follow-up question It is given that the sum of the roots of x2 – (3 – 4k)x – 6k = 0 is –9. (a) Find the value of k. (b) Find the product of the roots. (b) Product of roots = = –6(3) = –18 a c Product of roots =

If a and b are the roots of the quadratic equation x2 – 2x – 1 = 0, find the values of the following expression. (a) (a + 1)(b + 1) (b) a2 + b 2 a + b = = 2, ab = = -1 (a) (a + 1)(b + 1) = ab + b + a + 1 = ab + (a + b) + 1 = –1 + 2 + 1 = 2

If a and b are the roots of the quadratic equation x2 – 2x – 1 = 0, find the values of the following expression. (a) (a + 1)(b + 1) (b) a2 + b 2 a + b = = 2, ab = = -1 (b) a2 + b 2 = (a2 + 2ab + b 2) – 2ab = (a + b)2 – 2ab = (2)2 – (–1) = 5

Introduction to Complex Numbers

Imaginary Numbers Consider x2 = –1. x2 = –1 or 1 – = x ∵ and are not real numbers. ∴ The equation x2 = –1 has no real roots. x2 = –1 or 1 – = x They are called imaginary numbers. The square roots of negative numbers are called imaginary numbers. e.g. , , , 2 - 3

(c) For any positive real number p, (a) is denoted by i. 1 -  i.e. 1 - = i (b) 1 - = ·  i.e. i 2 = –1 (c) For any positive real number p, 1 - = · p  i.e. i p = - e.g. 4 - = 1 4 - · i 2 = = 9 - 1 9 - · 3i =

Complex Numbers A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and . 1 - i = a + b i Complex Number: Real part Imaginary part

Complex Numbers A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and . 1 - i = e.g. (i) 2 – i (ii) –3 (iii) 4i Real part: 2, imaginary part: –1 Real part: –3, imaginary part: 0 Real part: 0, imaginary part: 4

Complex Numbers A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and . 1 - i = e.g. (i) 2 – i (ii) –3 (iii) 4i Real part: 2, imaginary part: –1 Real part: –3, imaginary part: 0 Real part: 0, imaginary part: 4 For a complex number a + bi, if b = 0, then a + bi is a real number.

Complex Numbers A complex number is a number that can be written in the form a + bi, where a and b are real numbers, and . 1 - i = e.g. (i) 2 – i (ii) –3 (iii) 4i Real part: 2, imaginary part: –1 Real part: –3, imaginary part: 0 Real part: 0, imaginary part: 4 For a complex number a + bi, if a = 0 and b ≠ 0, then a + bi is an imaginary number.

Complex Number System Complex numbers Real numbers e.g. –3, 0 (i) Imaginary numbers e.g. 4i, –2i (ii) Sum of a non-zero real number and an imaginary number e.g. 2 – i, 1 + 5i

Follow-up question It is given that z = (k – 3) + (k + 1)i. If the imaginary part of z is 4, find the value of k, is z an imaginary number? (a) ∵ Imaginary part of z = 4 ∴ k + 1 = 4 k = 3

Follow-up question It is given that z = (k – 3) + (k + 1)i. If the imaginary part of z is 4, find the value of k, is z an imaginary number? (b) ∵ Real part = 3 – 3 = 0 Imaginary part ≠ 0 ∴ z is an imaginary number.

& a = c a + bi = c + di b = d Equality of Complex Numbers Two complex numbers (a + bi and c + di) are equal when both their and imaginary real parts parts are equal. a = c & a + bi = c + di Equality b = d

If x – 3i = yi, find the values of the real numbers x and y. ∵ The real parts are equal. ∴ x = 0 ∵ The imaginary parts are equal. ∴ y = –3

Follow-up question Find the values of the real numbers x and y if 2x + 4i = –8 + (y + 1)i. 2x + 4i = –8 + (y + 1)i By comparing the real parts, we have – 8 2 = x 4 – = x By comparing the imaginary parts, we have 1 4 + = y 3 = y

Operations of Complex Numbers

Let a + bi and c + di be two complex numbers. Addition (a + bi) + (c + di) = (a + c) + (b + d)i e.g. (1 + 2i) + (2 – i) = 1 + 2i + 2 – i = (1 + 2) + (2 – 1)i = 3 + i

Subtraction (a + bi) – (c + di) = (a – c) + (b – d)i e.g. (1 + 2i) – (2 – i) = 1 + 2i – 2 + i = (1 – 2) + (2 + 1)i = –1 + 3i

Follow-up question Simplify and express each of the following in the form a + bi. (a) (4 – 2i) + (3 + i) (b) (–5 + 3i) – (1 + 2i) (a) (4 – 2i) + (3 + i) = 4 – 2i + 3 + i = (4 + 3) + (–2 + 1)i = 7 – i (b) (–5 + 3i) – (1 + 2i) = –5 + 3i – 1 – 2i = (–5 – 1) + (3 – 2)i = –6 + i

Multiplication (a + bi)(c + di) = = ac + bci + adi + bdi 2 = ac + bci + adi + bd(–1) = (ac – bd) + (bc + ad)i (a + bi)(c) + (a + bi)(di) e.g. (1 + 2i)(2 – i) = (1 + 2i)(2) + (1 + 2i)(–i) = 2 + 4i – i – 2i2 = 2 + 3i – 2(–1) = 4 + 3i

Division i + - = 2 1 di c bi a + = di c - e.g. i - + = 2 4 ) ( )( di c · 2 1 di c bi a + = · di c - e.g. i - + = 2 4 2 ) ( )( di c bi a - + = i - + = ) 1 ( 4 2 5 d c i ad bc bd ac 2 ) ( + - = i = 5 i d c ad bc bd ac 2 + - = i =

Follow-up question Simplify and express each of the following in the form a + bi. (a) (1 + 3i)(–2 + 2i) (b) (a) (1 + 3i)(–2 + 2i) = (1 + 3i)(–2) + (1 + 3i)(2i) = –2 – 6i + 2i + 6i2 = –2 – 4i + 6(–1) = –8 – 4i

(b) i - + = · 3 1 2 4 i - + = ) 3 ( 1 6 12 2 4 i - + = ) 1 ( 9 6 10 4 i - = 10 i - = 1