1. The Discriminant 4.7A Chapter 4 Quadratic and Polynomial Equations
MATHPOWERTM 11, WESTERN EDITION
4.7.1

2. will give the roots of the quadratic equation.
The Nature of the Roots
The quadratic formula
will give the roots of the quadratic equation.
From the quadratic formula, the radicand,
b2 - 4ac, will determine the Nature of the Roots.
By the nature of the roots, we mean:
whether the equation has real roots or not
if there are real roots, whether they
are different or equal
The expression b2 - 4ac is called the discriminant
of the equation ax2 + bx + c = 0 because it discriminates
among the three cases that can occur.
4.7.2

3. x2 + x - 6 = 0 The Nature of the Roots [cont’d] The value of the
discriminant is
greater than zero.
x = -3 or 2
There are two distinct
real roots.
4.7.3

4. The Nature of the Roots [cont’d]
x2 + 2x + 1 = 0
The value of the
discriminant is 0.
There are two
equal real roots.
x = -1 or - 1
4.7.4

5. The Nature of the Roots [cont’d]
2x2 - 3x + 8 = 0
The value of the
discriminant is
less than zero.
Imaginary roots
4.7.5

6. The Nature of the Roots [cont’d]
We can make the following conclusions:
If b2 - 4ac > 0, then there are two different real roots.
If b2 - 4ac = 0, then there are two equal real roots.
If b2 - 4ac < 0, then there are no real roots.
Determine the nature of the roots:
x2 - 2x + 3 = 0
4x2 - 12x + 9 = 0
b2 - 4ac = (-2)2 - 4(1)(3)
= -8
b2 - 4ac = (-12)2 - 4(4)(9)
= 0
There are two equal
real roots.
The equation has no
real roots.
4.7.6

7. Determine the value of k for which the equation x2 + kx + 4 = 0 has
a) equal roots b) two distinct real roots c) no real roots
a) For equal roots, b2 - 4ac = 0.
Therefore, k2 - 4(1)(4) = 0
k = 0
k2 = 16
k = + 4
The equation has equal roots
when k = 4 and k = -4.
b) For two different real roots, b2 - 4ac > 0.
k > 0
k2 > 16
Therefore, k > 4 or k< -4. This may be written as | k | > 4.
Which numbers when squared
produce a result that is greater
than 16?
Which numbers when squared
produce a result that is less
than 16?
c) For no real roots, b2 - 4ac < 0.
k2 < 16
Therefore, -4 < k < 4. This may be written as | k | < 4.
4.7.7

8. Assignment Suggested Questions: Page 189 1, 2, 3, 18, 22,
28-30, 33, 34, 38
Read page 154, example 5.
Page , 51
4.7.8

9. Quadratic Equations - Word Problems
Chapter 4 Quadratic and Polynomial Equations
4.7B
Quadratic Equations -
Word Problems
MATHPOWERTM 11, WESTERN EDITION
4.7B.9

10. 1. The product of two consecutive even integers is 224.
Find the integers.
Let x = 1st even integer
Let y = 2nd even integer
y = x + 2
xy = 224
x(x + 2) = 224
x2 + 2x = 224
x2 + 2x = 0
(x + 16)(x - 14) = 0
x = - 16 or x = 14
When x = -16, the
integers are -16 and -14.
When x = 14, the
integers are 14 and 16.
Note there are two possible
solutions for x. Therefore, there
are two sets of answers.
4.7B.10

11. 2. The side of one square is 3 cm longer than the side of
another. If their combined area is 117 cm2, find
the dimensions of each square.
y = x + 3
Let x = side of the smaller square.
Let y = side of the larger square.
x2 + y2 = 117
Therefore, the smaller
square is 6 cm x 6 cm.
x2 + (x + 3)2 = 117
x2 + x2 + 6x + 9 =117
2x2 + 6x + 9 = 117
2x2 + 6x = 0
2(x2 + 3x - 54) = 0
2(x + 9)( x - 6) = 0
x = or x = 6
The larger square
is 9 cm x 9 cm.
Not possible
4.7B.11

12. 3. A factory is to be built on a lot that measures 80 m by 60 m.
A lawn of uniform width and equal in area to the factory, must
surround the factory. How wide is the strip of lawn, and
what are the dimensions of the factory?
Let x = the width of the strip. x
Area of the factory: x x
60 - 2x 60
2400 = (80 - 2x)(60 - 2x)
2400 = x + 4x2
0 = 4x x
0 = 4(x2 - 70x + 600)
0 = 4(x - 60)(x - 10)
x = or x = 10
80 - 2x x 80
Total area = 80 x 60
= 4800 m2
Area of the factory:
Therefore, the strip is 10 m wide, and
the factory is 60 m x 40 m.
= 2400 m2
4.7B.12

13. + = 5 4. Kelly drives 230 km in 5 h from her home in Whitehorse.
For the last 150 km of the trip, she increases her average speed
by 10 km/h. What is the average speed for each distance traveled?
d s t
d = st
Slower 80 x
Let x = slower speed
x + 10
150
Faster
80x x = 5x2 + 50x
5x x = 0
5(x2 - 36x - 160) = 0
5(x - 40)(x + 4) = 0
x = 40 or x = -4
Slower time + Faster time = Total time
x(x + 10) +
= 5
80(x + 10) + 150x = 5(x)(x + 10)
Therefore, the slower speed
was 40 km/h and the faster
was 50 km/h.
4.7B.13