Chapter V

The discriminant gives additional information on the nature of the roots beyond

simply whether there are any repeated roots: it also gives information on whether the

roots are real or complex, and rational or irrational. More formally, it gives information

on whether the roots are in the field over which the polynomial is defined, or are in an

extension field, and hence whether the polynomial factors over the field of coefficients.

This is most transparent and easily stated for quadratic and cubic polynomials; for

polynomials of degree 4 or higher this is more difficult to state.

TARGET SKILLS:

At the end of this chapter, students are expected to:

• determine discriminant, roots and coefficient;

• discuss the relation the roots and coefficient;

• find the sum and product of the roots; and

• change quadratic equation to discriminant formula.

Lesson 13

The Discriminant and the roots of a

Quadratic EquationOBJECTIVES:

At the end of this lesson, students are expected to:

determine discriminant and the roots;

compare discriminant and the nature of the roots; and

change quadratic equation to discriminant using the nature

of the roots.

Example

1. Find the x-intercept of y = 3x² - 6x + 4.

Solution: As already mentioned, the values of x for which 3x² - 6x + 4 = 0 give the x-

intercepts of the function. We apply the quadratic formula in solving the equation.

3x² - 6x + 4 = 0

x = −(ˉ 6)±√(ˉ 6)−4 (3)(4)

2(3) =

6±√ˉ 126

Since √ˉ 12 is not a real number, the equation 3x² - 6x + 4 = 0 has no real root.

This means that the parabola y = 3x² - 6x + 4 does not intersect the x-axis.

Let us write the equation in the form y = a(x – h)²+ k.

y = 3(x² – 2x)² + 4

= 3(x – 1)² + 1

∆ = b² - 4ac Roots of ax² + bx + c = 0

Positive Real and distinct

r = −b−√∆2a

s = −b+√∆2a

Zero Real and equal

r = s = ˉ b2a

Negative No real roots

Example 2. Use the disciminant to determine the nature of the roots of the

following quadratic equation.

a. x² - x + ¼ = 0

a = 1, b = ˉ1, c = ¼

b² - 4ac = (ˉ1)² - 4 (1)(¼)

= 1 – 1

= 0

There is only one solution, that is, a double root.

Note that x² - x = ½ = (x - ½), so that double root is ˉb/2a = ½.

b. 5x² - 4x + 1 = 0

a = 5, b = ˉ4, c = 1

b² - 4ac = (ˉ4)² - 4 (5)(1)

= 16 – 20

= ˉ4 < 0

There are no real roots since a negative number has no real square root.

Exercises:

Solve each by using the discriminant.

1. x² + 3x - ½ = 0

2. 3x² - 5x – 7 = 0

3. 6x² + 3x + 8 = 0

4. x² + 9x – 6 = 0

5. 8x² - 12x + 4 = 0

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Use the Discriminant to determine the nature of the root of the following Quadratic Equations.

1. x2- 2x – 3=0

_____________________________________________________

2. 6x2 – x – 1 = 0

_____________________________________________________

3. 2x2 – 50 = 0

_____________________________________________________

4. x2 – 8x + 12 = 0

_____________________________________________________

5. x2 + 5x – 14 = 0

_____________________________________________________

6. -4x2 – 4x + 1 = 0

_____________________________________________________

7. 7x2 + 2x – 1 = 0

_____________________________________________________

8. x2 + 3x = 40

_____________________________________________________

9. 3x2= 5x – 1

_____________________________________________________

10. 3x2+ 12 – 1=0

_____________________________________________________

11. (x-2)(x-3) = 4

_____________________________________________________

12.2x2 + 2x + 1 = 0

_____________________________________________________

13. 7x2 + 3 – 6x = 0

_____________________________________________________

14. 5x2 – 6x + 4 = 0

_____________________________________________________

15. 3x2 + 2x + 2 = 0

_____________________________________________________

Lesson 14

Relation between roots and coefficient

OBJECTIVES:

At the end of this lesson, students are expected to:

classify roots and coefficient;

discuss relations between the roots and coefficient of the quadratic

equation; and

find the sum and product of the roots of a given quadratic equation.

There are some interesting relations between the sum and the product of the roots of a quadratic equation. To discover these, consider the quadratic equation ax2 + bx + c = 0, where a ≠ 0.

Multiply both sides of this equation by 1/a so that the coefficient of x2 is 1.

1a

(ax2 + bx + c) = 1(0)a

We obtain an equivalent quadratic equation in the form

x2 + bxa

+ ca

= 0

If r and s are the roots of the quadratic equation ax2 + bx + c = 0, then from the

quadratic formula

r = −b+√b2−4 ac2a

and s = −b−√b2−4ac2a

Adding the roots, we obtain

r + s = −b+√b2−4 ac2a

+ −b−√b2−4ac2a

= ˉ 2b2a

= ˉ ba

Multiplying the roots, we obtain

rs = ¿¿ ¿¿

−b−(b ²−4ac )4 a ²

= ca

Observe the coefficient in the quadratic equation x2 + bx/a + c/a = 0. How do they

compare with the sum and the product of the roots? Did you observe the following?

1. The sum of the roots is equal to the negative of the coefficient of x.

r + s = ˉ ba

2. The product of the roots is equal to the constant term

rs = ca

An alternate way of arriving at these relations is as follows

Let r and s be the roots of x2 + bxa

+ ca

= 0. Then

(x - r)(x – s) = 0

Expanding gives, x2 – rx – sx + rs = 0

or x2 – (r + s)x + rs = 0

Comparing the coefficients of the corresponding terms, we obtain

r + s = ˉ ba

and rs = ca

The above relations between the roots and the coefficients provide a fast and

convenient means of checking the solutions of a quadratic equation.

Example: Solve and check. 2x2 + x – 6 = 0

Solutions: 2x2 + x – 6 = (2x – 3)(x + 2) = 0

x = 32

or x = -2

The roots are 32

and -2.

To check, we add the roots, 32

= (-2) = ˉ 12

= ˉ ba

and multiply them. 32

(-2) = -3 = ca

Example: Find the sum and the product of the roots of 3x2 – 6x + 8 = 0 without having to

first determine the roots.

Solution: The sum of the roots is r + s = ˉ ba

= ˉ (ˉ 6)3

= 2;

and their product is rs = ca

= 83

.

Exercises:

Without solving for the roots, find the sum and product of the roots of the following:

a. x2 – 5x – 36 = 0

b. 3x2 + 2x – 1 = 0

c. 5x2 + x – 1 =0

d.2x+x=3

e. x (x−2 )=(2 x+1 ) ( x+1 )

f. 3px2 – 4pqx + x2 = 0

Name: ___________________ Section: _______

Instructor: ________________ Date: _______ Rating: ____

Instruction: Without solving the roots, find the sum and product of the roots of the following.

1. 6x2 – 5x + 2 = 0

_____________________________________________________

2. x2 + x – 182 = 0

_____________________________________________________

3. x2 – 5x – 14 = 0

_____________________________________________________

4. 2x2 – 9x + 8 = 0

_____________________________________________________

5. 3x2 - 5x – 2 = 0

_____________________________________________________

6. X2 – 8x – 9 = 0

_____________________________________________________

7. 2x2 – 3x – 9 = 0

_____________________________________________________

8. X2 + x – 2 - √2=0

_____________________________________________________

9. 3x2 + 2x – 8 = 0

_____________________________________________________

10.16x2 – 24x + 12=0

_____________________________________________________

11.x2 – 6x + 25 = 0

_____________________________________________________

12.3x2 + x – 2 = 0

_____________________________________________________

13.5x2 + 11x – 8 = 0

_____________________________________________________

14.x2 – 8x + 16 = 0

_____________________________________________________

15.4x2 – 16x + 10 = 0

_____________________________________________________

A. Use the discriminant to determine which of the following quadratic equations

have two, one or no real roots. Give reasons.

1. x2 – 5x – 5 = 0

2. x2 – 3x – 2 = 3x – 11

3. 5x2 – 9x = 2x – 7

4. 2x2 – 7x + 8 = 0

5. 3 – 4x – 2x2 = 0

6. 4x2 – 9x + 5 = 3x – 7

B. By using the relations between roots and coefficients, determine if the given #s

are roots of the corresponding given equation.

1. 6x2 – 5x + 3 = 0 (1/6, -1)

2. x2 + x _ 182 = 0 (13, -4)

3. x2 – 5x – 14 =0 (2, -7)

4. 2x – 9x + 8 = 0 (3, 4/3)

5. 3x2 – 5x – 2 = 0 (2, -1/3)

C. Given one roots of the equation, find the other.

1. x2 – 8x – 9 = 0; r= 1

2. 2x2 – 3x – 9 = 0; r= 3

3. x2 + x – 2 - √2 = 0; r= √2