SCHOLAR Study Guide
National 5 Mathematics
Course Materials
Topic 19:
Solving
equations
Authored by:
Margaret Ferguson
Reviewed by:
Jillian Hornby
Previously authored by:
Eddie Mullan
Heriot-Watt University
Edinburgh EH14 4AS, United Kingdom.
quadratic
First published 2014 by Heriot-Watt University.
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SCHOLAR Study Guide Course Materials Topic 19: National 5 Mathematics
1. National 5 Mathematics Course Code: C747 75
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1
Topic 1
Solving quadratic equations
Contents
19.1 Solving quadratic equations graphically . . . . . . . . . . . . . . . . . . . . . .
3
19.2 Solving quadratic equations by factorising . . . . . . . . . . . . . . . . . . . .
19.3 Solving quadratic equations using the quadratic formula . . . . . . . . . . . .
7
9
19.4 Determining and interpreting the nature of the roots using the discriminant . .
19.5 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
17
19.6 End of topic test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
2
TOPIC 1. SOLVING QUADRATIC EQUATIONS
Learning objectives
By the end of this topic, you should be able to:
•
•
solve a quadratic equation:
◦
graphically;
◦
by factorising;
◦
using the quadratic formula;
identify and interpret the nature of the roots of a quadratic using the discriminant.
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
1.1
3
Solving quadratic equations graphically
Quadratic equations are solved to find the value or values of x where the graph crosses
the x-axis.
We already know from Topic 18: Identifying the features of a quadratic function how to
identify the coordinates of the zeros or roots of a quadratic. Work through the following
activity to remind yourself.
Solving quadratic equations graphically
Given a graph of a quadratic function you should be able to find the solutions of the
function, that is, the values of x when y = 0. Use the following examples to see how
this is done.
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4
TOPIC 1. SOLVING QUADRATIC EQUATIONS
..........................................
Key point
The solutions of the equation ax 2 + bx + c = 0 will be the x-coordinates of the
points where the graph of y = ax 2 + bx + c cuts the x-axis.
Example
Problem:
The diagram shows the graph of the function y = x 2 + 2x − 8.
Use the graph to solve the equation x 2 + 2x − 8 = 0.
Solution:
The curve cuts the x-axis at - 4 and 2.
So the solutions to the equation are x = − 4 and x = 2.
..........................................
Q1:
The diagram shows the graph of the function y = x 2 + 2x − 3.
Use the graph to solve the equation x 2 + 2x − 3 = 0.
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
5
..........................................
Solving quadratic equations graphically practice
Q2:
The diagram shows the graph of the function y = 9 − x 2 .
Use the graph to solve the equation 9 − x 2 = 0.
..........................................
Q3:
The diagram shows the graph of the function y = 2x 2 + 2x − 7 · 5.
Use the graph to solve the equation 2x 2 + 2x − 7 · 5 = 0.
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Go online
6
TOPIC 1. SOLVING QUADRATIC EQUATIONS
..........................................
Solving quadratic equations graphically exercise
Q4:
Go online
The diagram shows the graph of the function y = x 2 − x − 6.
Use the graph to solve the equation x 2 − x − 6 = 0.
..........................................
Q5:
The diagram shows the graph of the function y = 12 + x − x 2 .
Use the graph to solve the equation 12 + x − x 2 = 0.
..........................................
Q6:
The diagram shows the graph of the function y = 2x 2 − 11x + 9.
Use the graph to solve the equation 2x 2 − 11x + 9 = 0.
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
7
..........................................
1.2
Solving quadratic equations by factorising
We already know from Topic 9: Factorising how to factorise a quadratic and from Topic
18: Identifying the features of a quadratic function how to identify the roots or zeros.
Work through the following activity to remind yourself.
Solving quadratic equations by factorisation
x2 + 3x + 2 is a simple quadratic expression.
It can be factorised as (x + 2)(x + 1).
The coefficient of x is 3 . . . this comes from 2 + 1.
The constant term is 2 . . . this comes from 2 × 1.
Key point
The general factorised form is (x + m)(x + n). . .
. . . where the coefficient of x is m + n
. . . and the constant term is m × n.
..........................................
Examples
1.
Problem:
Factorise x2 + 4x − 12.
Solution:
Step 1: List the pairs of numbers that multiply to make -12.
Step 2: Work out the sum of each pair.
Step 3: Pick the pair that sum to + 4 which is -2 and +6.
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8
TOPIC 1. SOLVING QUADRATIC EQUATIONS
Step 4: Write down the factors of the expression: (x − 2)(x + 6).
..........................................
2.
Problem:
Solve x2 + 4x − 12 = 0.
Solution:
Step 1:
First factorise the equation to give (x − 2)(x + 6) = 0.
Step 2:
(x − 2)(x + 6) = 0 will be true if either bracket is equal to zero.
(x − 2) = 0 or (x + 6) = 0
Step 3:
Solve x − 2 = 0 or x + 6 = 0 to give x = 2 or x = − 6.
Therefore the solution to x 2 + 4x − 12 = 0 is x = 2 or x = − 6.
..........................................
Solving quadratic equations by factorisation practice
Go online
Q7: Factorise x2 + 2x − 15.
(Remember: You can check whether your answer is correct by multiplying out the
brackets. If you are correct, the result should give the original quadratic.)
..........................................
Q8:
Solve x2 + x − 6 = 0.
..........................................
Q9:
Solve 2x2 + 9x − 18 = 0.
Remember as we now have 2x 2 finding the factorised form requires more thought.
..........................................
Q10: Solve 1 − 2x + x2 = 0.
..........................................
Solving quadratic equations by factorisation exercise
Q11: Factorise x2 − 2x − 24.
Go online
..........................................
Q12: Use factorisation to solve the equation x 2 − 5x + 6 = 0.
a) Factorise x2 − 5x + 6.
b) Solve x2 − 5x + 6 = 0.
..........................................
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
9
Q13: Solve x2 − 25 = 0.
..........................................
Q14: Solve x2 + 6x = 0.
..........................................
Q15: Solve 3 − 2x − x2 = 0.
..........................................
Q16: Solve 4x2 + 8x + 3 = 0.
..........................................
1.3
Solving quadratic equations using the quadratic formula
Sometimes it is not possible to factorise the quadratic and the graph has not been
provided. In this situation you need the quadratic formula to be able to solve the
quadratic equation.
Key point
The Quadratic Formula
For a quadratic equation of the form ax 2 + bx + c = 0.
x =
−b ±
√
b 2 − 4ac
2a
i.e. the solutions to the equation ax 2 + bx + c = 0 are: x
x =
−b −
√
b 2−
2a
4ac
=
−b +
√
b 2 − 4ac
2a
and
.
Top tip
The Quadratic Formula is on the National 5 Formula Sheet which will be given to
you in the exam.
Usually, the quadratic formula would be used to solve ax 2 + bx + c = 0 if it
cannot be easily factorised but a big hint is given when you are asked to solve the
quadratic equation to a number of decimal places or significant figures.
Example
Problem:
Solve x2 + 3x − 5 = 0. Give your answers to 3 significant figures.
Solution:
Comparing the equation x 2 + 3x − 5 = 0 with the standard one of ax 2 + bx + c = 0
we see that a = 1, b = 3 and c = − 5.
Substituting these values into the quadratic formula gives:
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10
TOPIC 1. SOLVING QUADRATIC EQUATIONS
√
b2 − 4ac
2a
32 − 4 × 1 × (−5)
−3 ±
⇒ x =
2 × 1
√
−3 ± 9 + 20
=
2
√
√
−3 − 29
−3 + 29
or x =
⇒ x =
2
2
⇒ x = 1 · 19 and x = −4 · 19 (to 3 s.f.)
..........................................
x =
−b ±
Top tip
If after substituting into the Quadratic Formula you find you have a negative value
underneath the square root you have made a mistake.
Go back and check the steps in your working. It is not possible to find the square
root of a negative number.
Solving quadratic equations using the quadratic formula practice
Q17: Solve x2 + 2x − 7 = 0.
Go online
..........................................
Q18: Solve 2x2 − 5x + 1 = 0 giving your solutions correct to 1 d.p.
..........................................
Solving quadratic equations using the quadratic formula exercise
Q19: Solve x2 − x − 4 = 0 using the quadratic formula.
Go online
..........................................
Q20: Solve x2 − 6x − 8 = 0 using the quadratic formula, give your answer correct to
1 d.p.
..........................................
Q21: Solve 2x2 − 2x − 4 = 0 using the quadratic formula, give your answer correct
to 1 d.p.
..........................................
1.4
Determining and interpreting the nature of the roots
using the discriminant
We already know from Topic 18: Identifying the features of a quadratic function how to
identify the shape of a quadratic graph but this does not tell us its position with respect
to the x-axis. The discriminant will help us to determine this.
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
11
Key point
The discriminant of the quadratic equation ax 2 + bx + c = 0 is b2 − 4ac.
Notice b2 − 4ac is the expression underneath the square root in the Quadratic Formula.
Key point
The following conditions on the discriminant hold:
•
If b2 − 4ac < 0, there are no real roots.
•
If b2 − 4ac = 0, the roots are real and equal.
•
If b2 − 4ac > 0, the roots are real and distinct.
These conditions can be related to the position of the graph of a quadratic with respect
to the x-axis.
Roots of an equation
Go online
No Real Roots
The discriminant is less than zero.
The graph never crosses the x-axis.
Real and Equal Roots
The discriminant is equal to zero.
The graph only touches the x-axis at one
point.
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
Real and Distinct Roots
The discriminant is greater than zero.
The graph crosses the x-axis at two
distinct points.
..........................................
Here are some examples of both the graphs and the expressions in relation to the
discriminant.
Roots and the discriminant
An equation with no real roots has a discriminant with a negative value.
Go online
Therefore the graphs do not cross the x-axis.
An equation with real and equal roots has a discriminant equal to zero.
Therefore the graph touches the x-axis at only one point.
An equation with real and distinct roots has a discriminant with a positive value.
Therefore the graphs cross the x-axis at two distinct points.
..........................................
Examples
1. Real and distinct roots
Problem:
Determine the nature of the roots of the quadratic y = x 2 − 2x − 1
Solution:
From the equation of the quadratic we get a = 1, b = − 2 and c = − 1.
Using the discriminant this gives b2 − 4ac = (−2)2 − 4 × 1 × (−1) = 8
Since the discriminant is greater than zero the roots are real and distinct.
i.e. there are 2 solutions to x2 − 2x − 1 = 0
The shape of the graph is a smiley face and looks like this,
..........................................
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
2. Real and equal roots
Problem:
Determine the nature of the roots of the quadratic y = x 2 + 6x + 9
Solution:
From the equation of the quadratic we get a = 1, b = 6 and c = 9.
Using the discriminant this gives b2 − 4ac = 62 − 4 × 1 × 9 = 0
Since the discriminant is equal to zero the roots are real and equal.
i.e. there are really 2 solutions to x2 + 6x + 9 = 0 but they are the same so we think
of this quadratic as having 1 solution.
The shape of the graph is a smiley face and looks like this,
..........................................
3. No real roots
Problem:
Find the roots of the quadratic y = x 2 − x + 4
Solution:
From the equation of the quadratic we get a = 1, b = − 1 and c = 4.
Using the discriminant this gives b2 − 4ac = (−1)2 − 4 × 1 × 4 = − 15
Since the discriminant is less than zero there are no real roots.
i.e. there are no solutions to x2 − x + 4 = 0. This is because we cannot find the
square root of a negative number.
The shape of the graph is a smiley face and looks like this,
..........................................
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14
TOPIC 1. SOLVING QUADRATIC EQUATIONS
Determining and interpreting the nature of the roots using the
discriminant exercise
Q22:
Go online
Examine the discriminant of these quadratics, determine the nature of the roots and
state whether they are:
•
real and distinct
•
real and equal
•
no real roots
a) 7x2 − 3x − 8
b) 7x2 − 3x + 8
c) 2x2 − 8x + 8
d) x2 + 6x + 24
..........................................
Q23:
Identify the position of the following graphs. Do they touch; avoid; cross the x-axis?
•
real and distinct
•
real and equal
•
no real roots
a) x2 + 8x + 32
b) 7x2 − 3x − 4
c) 4x2 − 8x + 4
d) 7x2 − 3x + 4
..........................................
The discriminant can tell us more about the roots.
Example
Problem:
Prove that the roots of the equation 5x 2 + 2x − 7 = 0 are real and rational.
Solution:
There are two ways of proving this.
1. Factorise the quadratic equation.
5x2 + 2x − 7 = 0
(5x + 7) (x − 1) = 0
5x + 7 = 0 or
5x = −7 or
The solutions
− 75
x − 1 = 0
x = 1
and 1 are both rational numbers.
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
15
2. Find the discriminant.
b2 − 4ac = 22 − 4 × 5 × (−7)
= 4 + 140
= 144
Since 144 is a positive square number (i.e. 122 = 144) we know that the roots are real
and rational. This is because the discriminant is the part of the quadratic formula under
the square root.
If we were to find
√ the solutions using the quadratic formula we would get,
−2 ± 144
x =
2 × 5
−2 − 12
−2 + 12
or x =
x =
10
10
again we find that both solutions are rational.
..........................................
Note: It is quicker to calculate the discriminant to determine whether the roots are
rational or irrational.
Key point
If the discriminant is a positive square number then the roots are real and rational.
Example
Problem:
Prove that the roots of the equation 3x 2 − 8x + 2 = 0 are real and irrational.
Solution:
b2 − 4ac = (−8)2 − 4 × 3 × 2
= 64 − 24
= 40
Since 40 is positive but not a square number (i.e.
that the roots are real and irrational.
√
40 is an irrational number) we know
..........................................
Key point
If the discriminant is positive but not a square number then the roots are real and
irrational.
Rational and irrational roots practice
Q24: Prove that the roots of the equation 2x 2 + 3x − 1 = 0 are real and irrational.
..........................................
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16
TOPIC 1. SOLVING QUADRATIC EQUATIONS
Q25: Prove that the roots of the equation 3x 2 − x − 2 = 0 are real and rational.
..........................................
Rational and irrational roots exercise
Q26: Prove that the roots of the equation 6x 2 + 5x + 1 = 0 are real and rational.
Go online
..........................................
Q27: Prove that the roots of the equation 3x 2 + 6x − 4 = 0 are real and irrational.
..........................................
Q28: Are the roots of the equation 5x 2 + 6x − 3 = 0 real and rational or real and
irrational?
..........................................
Q29: Are the roots of the equation 2x 2 − 13x + 3 = 0 real and rational or real and
irrational?
..........................................
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
1.5
Learning points
Solving a quadratic equation graphically
•
The zeros or roots of a quadratic are the point(s) where the graph crosses the
x-axis and will take the form (p,0) and (q,0).
•
The solution is x = p and x = q.
Solving a quadratic equation by factorising
•
A quadratic equation of the form y = ax 2 + bx + c must be factorised to take
the form y = (x − p)(x − q).
•
The roots are (p,0) and (q,0).
•
To calculate the solution pull the brackets apart:
◦
•
x − p = 0 and x − q = 0.
The solution is x = p and x = q.
Solving a quadratic equation using the quadratic formula
•
From a quadratic equation of the form y = ax 2 + bx + c you must identify the
values of a, b and c.
•
Substitute a, b and c into the quadratic formula.
•
x =
•
Remember to split the expression into:
√
−b± b2 −4ac
2a
◦
x =
◦
x =
√
−b+ b2 −4ac
2a
√
−b− b2 −4ac
2a
and
Identifying and interpreting the nature of the roots of a quadratic using the
discriminant
•
From a quadratic equation of the form y = ax 2 + bx + c you must identify the
values of a, b and c.
•
The discriminant of a quadratic is b2 − 4ac.
•
If b2 − 4ac > 0 (positive), the roots are real and distinct.
◦
•
•
There are two solutions.
If b2 − 4ac = 0, the roots are real and equal.
◦
There is one solution.
◦
There are two really but they are both the same.
If b2 − 4ac < 0 (negative), there are no real roots.
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17
18
TOPIC 1. SOLVING QUADRATIC EQUATIONS
◦
There are no solutions.
•
If b2 − 4ac is positive and a square number, the roots are real and rational.
•
If b2 − 4ac is positive and not a square number, the roots are real and irrational.
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TOPIC 1. SOLVING QUADRATIC EQUATIONS
1.6
19
End of topic test
End of topic 19 test
Solving Quadratic Equations Graphically
Go online
Q30:
The diagram shows the graph of the function y = x 2 + 3x − 4.
Use the graph to solve the equation y = x 2 + 3x − 4.
..........................................
Q31:
The diagram shows the graph of the function y = 12 + x − x 2 .
Use the graph to solve the equation y = 12 + x − x 2 .
..........................................
Q32:
The diagram shows the graph of the function y = 2x 2 − 11x + 9.
Use the graph to solve the equation y = 2x 2 − 11x + 9.
© H ERIOT-WATT U NIVERSITY
20
TOPIC 1. SOLVING QUADRATIC EQUATIONS
..........................................
Solving Quadratic Equations by Factorising
Q33: Solve x2 + 6x = 0
..........................................
Q34: Solve x2 − 25 = 0
..........................................
Q35: Solve x2 − 2x − 15 = 0
..........................................
Q36: Solve 3 − 2x − x2 = 0
..........................................
Q37: Solve 4x2 + 8x + 3 = 0
..........................................
Solving Quadratic Equations using the Quadratic Formula
Q38: Solve x2 + 7x + 8 = 0 using the quadratic formula, give your answer correct to
1 d.p.
..........................................
Q39: Solve x2 − 6x − 8 = 0 using the quadratic formula, give your answer correct to
1 d.p.
..........................................
Q40: Solve x2 − 3x − 8 = 0 using the quadratic formula, give your answer correct to
1 d.p.
..........................................
Q41: Solve 2x2 − 2x − 4 = 0 using the quadratic formula, give your answer correct
to 1 d.p.
..........................................
© H ERIOT-WATT U NIVERSITY
TOPIC 1. SOLVING QUADRATIC EQUATIONS
The Discriminant
Q42:
Examine the discriminant of these quadratics, determine the nature of the roots and
state whether they are:
•
real and distinct
•
real and equal
•
no real roots
a) 4x2 − 8x + 4
b) 5x2 − 3x + 6
c) 5x2 − 3x − 6
d) x2 + 8x + 32
..........................................
Q43:
Examine the discriminant of these quadratics, determine the nature of the roots and
state whether they are:
•
real and distinct
•
real and equal
•
no real roots
a) 8x2 + 9x
b) 8x2 − 9
c) 8 − 8x + 2x2
d) 28 + 7x − x2
..........................................
Q44: Are the roots of the equation x 2 − 7x + 3 = 0 are real and rational or real and
irrational?
..........................................
..........................................
© H ERIOT-WATT U NIVERSITY
21
22
GLOSSARY
Glossary
discriminant
the discriminant of the quadratic equation ax 2 + bx + c = 0 is b2 − 4ac
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ANSWERS: TOPIC 19
Answers to questions and activities
19 Solving quadratic equations
Answers from page 4.
Q1:
The curve cuts the x-axis at -3 and 1.
The solutions to the equation are x = − 3 and x = 1.
Solving quadratic equations graphically practice (page 5)
Q2: x=-3,x=3
Q3: x=-2.5,x=1.5
Solving quadratic equations graphically exercise (page 6)
Q4: x = − 2 and x = 3
Q5: x = − 3 and x = 4
Q6: x = 1 and x = 4 · 5
Solving quadratic equations by factorisation practice (page 8)
Q7: (x + 5)(x − 3)
Q8: (x − 2)(x + 3) = 0
x − 2 = 0 or x + 3 = 0
x = 2 or x = − 3
Q9: (2x − 3)(x + 6) = 0
2x − 3 = 0 or x + 6 = 0
2x = 3 or x = − 6
x = 1·5
The solutions are x = 1 · 5 or x = − 6.
Q10: If we factorise 1 − 2x + x2 = 0 we get (1 − x)(1 − x) = 0 or (1 − x)2 = 0.
1 − x = 0 so the solution is x = 1.
This equation really has two solutions but since they are both the same we think of it as
having only one solution.
Solving quadratic equations by factorisation exercise (page 8)
Q11: (x − 6)(x + 4)
Q12:
© H ERIOT-WATT U NIVERSITY
23
24
ANSWERS: TOPIC 19
a) (x − 2)(x − 3)
b) x = 2 and x = 3
Q13:
Steps:
•
Factorise x2 − 25. (x − 5)(x + 5)
Answer: x = − 5 and x = 5
Q14:
Steps:
•
Factorise x2 + 6x. x(x + 6)
Answer: x = − 6 and x = 0
Q15:
Steps:
•
Factorise 3 − 2x − x2 . (x + 3)(−x + 1)
Answer: x = − 3 and x = 1
Q16:
Steps:
•
Factorise 4x2 + 8x + 3. (2x + 3)(2x + 1)
Answer: x = − 1 · 5 and x = − 0 · 5
Solving quadratic equations using the quadratic formula practice (page 10)
Q17:
From x2 + 2x − 7
c = −7
b = 2
we see that a = 1
√
−b ± b2 − 4ac
x =
2a
22 − 4 × 1 × (−7)
−2 ±
⇒ x =
2 × 1
√
−2 ± 4 + 28
=
2
√
√
−2 − 32
−2 + 32
or x =
⇒ x =
2
2
⇒ x = 1 · 8 and x = −3 · 8 (to 1 d.p.)
Q18:
we see that a = 2
From 2x2 − 5x + 1
b = −5
c = 1
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 19
x =
⇒ x =
−b ±
25
√
b2 − 4ac
2a − (−5) ±
(−5)2 − 4 × 2 × 1
2 × 2
25 − 8
=
√
√4
5 − 17
5 + 17
or x =
⇒ x =
4
4
⇒ x = 2 · 3 and x = 0 · 2 (to 1 d.p.)
5 ±
√
Solving quadratic equations using the quadratic formula exercise (page 10)
Q19:
Steps:
•
To solve the equation x 2 − x − 4 = 0 you must substitute a = 1, b = − 1 and
c = − 4 into the quadratic formula.
•
To evaluate the quadratic formula, you must start by calculating b 2 − 4ac. 17
Answer: x = − 1 · 6 and x = 2 · 6
Q20: x = − 1 · 1 and x = 7 · 1
Q21: x = − 1 and x = 2
Determining and interpreting the nature of the roots using the discriminant
exercise (page 14)
Q22:
a) real and distinct; b2 − 4ac = 233 which is positive.
b) no real roots; b2 − 4ac = − 215 which is negative.
c) real and equal; b2 − 4ac = 0.
d) no real roots; b2 − 4ac = − 60 which is negative.
Q23:
a) no real roots; b2 − 4ac = − 64 so the graph will avoid the x-axis.
b) real and distinct; b2 − 4ac = 121 so the graph will cross the x-axis.
c) real and equal; b2 − 4ac = 0 so the graph will only touch the x-axis.
d) no real roots; b2 − 4ac = − 103 so the graph will avoid the x-axis.
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ANSWERS: TOPIC 19
Rational and irrational roots practice (page 15)
Q24:
b2 − 4ac = 32 − 4 × 2 × (−1)
= 9 + 8
= 17
Since 17 is positive but not a square number (i.e.
that the roots are real and irrational.
√
17 is an irrational number) we know
Q25:
b2 − 4ac = (−1)2 − 4 × 3 × (−2)
= 1 + 24
= 25
Since 25 is positive and a square number (i.e. 5 2 = 25) we know that the roots are real
and rational.
Rational and irrational roots exercise (page 16)
Q26:
b2 − 4ac = 52 − 4 × 6 × 1
= 25 − 24
= 1
Since 1 is positive and a square number (i.e. 1 2 = 1) we know that the roots are real
and rational.
Q27:
b2 − 4ac = 62 − 4 × 3 × (−4)
= 36 + 48
= 84
Since 84 is positive but not a square number (i.e.
that the roots are real and irrational.
√
84 is an irrational number) we know
Q28:
Steps:
•
What is the value of the discriminant? 96
•
Is the discriminant positive? yes
•
Is the discriminant a square number? yes
•
Use this to determine the answer.
Answer: real and rational
Q29:
Steps:
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 19
•
What is the value of the discriminant? 193
•
Is the discriminant positive? yes
•
Is the discriminant a square number? no
•
Use this to determine the answer.
Answer: real and irrational
End of topic 19 test (page 19)
Q30: x = 1 and x = − 4
Q31: x = − 3 and x = 4
Q32: x = 1 and x = 4 · 5
Q33:
Steps:
•
Factorise x2 + 6x. x(x + 6)
Answer: x = − 6 and x = 0
Q34:
Steps:
•
Factorise x2 − 25 = 0. (x − 5)(x + 5)
Answer: x = − 5 and x = 5
Q35:
Steps:
•
Factorise x2 − 2x − 15. (x − 5)(x + 3)
Answer: x = − 3 and x = 5
Q36:
Steps:
•
Factorise 3 − 2x − x2 . (x + 3)(−x + 1)
Answer: x = − 3 and x = 1
Q37:
Steps:
•
Factorise 4x2 + 8x + 3. (2x + 3)(2x + 1)
Answer: x = − 0 · 5 and x = − 1 · 5
Q38:
Hint:
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28
ANSWERS: TOPIC 19
•
To solve the equation x 2 + 7x + 8 = 0, you must substitute a = 1, b = 7 and
c = 8 into the quadratic formula.
Steps:
•
To evaluate the quadratic formula, begin by calculating b 2 − 4ac. b2 − 4ac = 17
Answer: x = − 5 · 6 and x = − 1 · 4
Q39:
Hint:
•
To solve the equation x 2 − 6x − 8 = 0, you must substitute a = 1, b = − 6
and c = − 8 into the quadratic formula.
Steps:
•
To evaluate the quadratic formula, begin by calculating b 2 − 4ac. b2 − 4ac = 68
Answer: x = − 1 · 1 and x = 7 · 1
Q40:
Hint:
•
To solve the equation x 2 − 3x − 8 = 0, you must substitute a = 1, b = − 3
and c = − 8 into the quadratic formula.
Steps:
•
To evaluate the quadratic formula, begin by calculating b 2 − 4ac. b2 − 4ac = 41
Answer: x = − 1 · 7 and x = 4 · 7
Q41:
Hint:
•
To solve the equation 2x 2 − 2x − 4 = 0, you must substitute a = 2, b = − 2
and c = − 4 into the quadratic formula.
Steps:
•
To evaluate the quadratic formula, begin by calculating b 2 − 4ac. b2 − 4ac = 36
Answer: x = − 1 and x = 2
Q42:
a) real and equal; b 2 − 4ac = 0.
b) no real roots; b2 − 4ac = − 111 which is negative.
c) real and distinct; b2 − 4ac = 129 which is positive.
d) no real roots; b2 − 4ac = − 64 which is negative.
Q43:
a) real and distinct; b2 − 4ac = 81 which is positive.
© H ERIOT-WATT U NIVERSITY
ANSWERS: TOPIC 19
b) no real roots; b2 − 4ac = − 288 which is negative.
c) real and equal; b2 − 4ac = 0.
d) real and distinct; b2 − 4ac = 161 which is positive.
Q44: real and irrational
Steps:
•
What is the value of the discriminant? 37
•
Is the discriminant positive? yes
•
Is the discriminant a square number? no
•
Use this to determine the answer.
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