SCHOLAR Study Guide National 5 Mathematics Course ... · SCHOLAR Study Guide National 5 Mathematics Course Materials Topic 19: Solving quadratic equations Authored by: Margaret Ferguson

SCHOLAR Study Guide

National 5 Mathematics

Course MaterialsTopic 19: Solving quadraticequations

Authored by:Margaret Ferguson

Reviewed by:Jillian Hornby

Previously authored by:Eddie Mullan

Heriot-Watt University

Edinburgh EH14 4AS, United Kingdom.

First published 2014 by Heriot-Watt University.

This edition published in 2016 by Heriot-Watt University SCHOLAR.

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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 . . . . . . . . . . . . . . . . . . . . 7

19.3 Solving quadratic equations using the quadratic formula . . . . . . . . . . . . 9

19.4 Determining and interpreting the nature of the roots using the discriminant . . 10

19.5 Learning points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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.

© HERIOT-WATT UNIVERSITY

TOPIC 1. SOLVING QUADRATIC EQUATIONS 3

1.1 Solving quadratic equations graphically

Quadratic equations are solved to find the value or values of x where the graph crossesthe x-axis.

We already know from Topic 18: Identifying the features of a quadratic function how toidentify the coordinates of the zeros or roots of a quadratic. Work through the followingactivity to remind yourself.

Solving quadratic equations graphically

Go online

Given a graph of a quadratic function you should be able to find the solutions of thefunction, that is, the values of x when y = 0. Use the following examples to see howthis is done.



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Key point

The solutions of the equation ax2 + bx + c = 0 will be the x-coordinates of thepoints where the graph of y = ax2 + bx + c cuts the x-axis.

Example

Problem:

The diagram shows the graph of the function y = x2 + 2x − 8.Use the graph to solve the equation x2 + 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.

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Q1:

The diagram shows the graph of the function y = x2 + 2x − 3.Use the graph to solve the equation x2 + 2x − 3 = 0.



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Solving quadratic equations graphically practice

Go online

Q2:

The diagram shows the graph of the function y = 9 − x2.Use the graph to solve the equation 9 − x2 = 0.

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Q3:

The diagram shows the graph of the function y = 2x2 + 2x − 7 · 5.Use the graph to solve the equation 2x2 + 2x − 7 · 5 = 0.



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Solving quadratic equations graphically exercise

Go online

Q4:

The diagram shows the graph of the function y = x2 − x − 6.Use the graph to solve the equation x2 − x − 6 = 0.

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Q5:

The diagram shows the graph of the function y = 12 + x − x2.Use the graph to solve the equation 12 + x − x2 = 0.

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Q6:

The diagram shows the graph of the function y = 2x2 − 11x + 9.Use the graph to solve the equation 2x2 − 11x + 9 = 0.



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1.2 Solving quadratic equations by factorising

We already know from Topic 9: Factorising how to factorise a quadratic and from Topic18: 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

Go online

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.

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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.



Step 4: Write down the factors of the expression: (x − 2)(x + 6).

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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 x2 + 4x − 12 = 0 is x = 2 or x = − 6.

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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 thebrackets. If you are correct, the result should give the original quadratic.)

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Q8: Solve x2 + x − 6 = 0.

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Q9: Solve 2x2 + 9x − 18 = 0.

Remember as we now have 2x2 finding the factorised form requires more thought.

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Q10: Solve 1 − 2x + x2 = 0.

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Solving quadratic equations by factorisation exercise

Go online

Q11: Factorise x2 − 2x − 24.

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Q12: Use factorisation to solve the equation x2 − 5x + 6 = 0.

a) Factorise x2 − 5x + 6.

b) Solve x2 − 5x + 6 = 0.

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Q13: Solve x2 − 25 = 0.

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Q14: Solve x2 + 6x = 0.

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Q15: Solve 3 − 2x − x2 = 0.

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Q16: Solve 4x2 + 8x + 3 = 0.

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1.3 Solving quadratic equations using the quadratic formula

Sometimes it is not possible to factorise the quadratic and the graph has not beenprovided. In this situation you need the quadratic formula to be able to solve thequadratic equation.

Key point

The Quadratic Formula

For a quadratic equation of the form ax2 + bx + c = 0.

x = −b ± √b 2− 4ac2a

i.e. the solutions to the equation ax2 + bx + c = 0 are: x = −b +√b 2− 4ac2a and

x = −b − √b 2− 4ac2a .

Top tip

The Quadratic Formula is on the National 5 Formula Sheet which will be given toyou in the exam.Usually, the quadratic formula would be used to solve ax2 + bx + c = 0 if itcannot be easily factorised but a big hint is given when you are asked to solve thequadratic 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 x2 + 3x − 5 = 0 with the standard one of ax2 + bx + c = 0we see that a = 1, b = 3 and c = − 5.Substituting these values into the quadratic formula gives:



x =−b ± √

b2 − 4ac

2a

⇒ x =−3 ± √

32 − 4 × 1 × (−5)

2 × 1

=−3 ± √

9 + 20

2

⇒ x =−3 +

√29

2or x =

−3 − √29

2⇒ x = 1 · 19 and x = −4 · 19 (to 3 s.f.)

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Top tip

If after substituting into the Quadratic Formula you find you have a negative valueunderneath the square root you have made a mistake.Go back and check the steps in your working. It is not possible to find the squareroot of a negative number.

Solving quadratic equations using the quadratic formula practice

Go online

Q17: Solve x2 + 2x − 7 = 0.

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Q18: Solve 2x2 − 5x + 1 = 0 giving your solutions correct to 1 d.p.

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Solving quadratic equations using the quadratic formula exercise

Go online

Q19: Solve x2 − x − 4 = 0 using the quadratic formula.

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Q20: Solve x2 − 6x − 8 = 0 using the quadratic formula, give your answer correct to1 d.p.

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Q21: Solve 2x2 − 2x − 4 = 0 using the quadratic formula, give your answer correctto 1 d.p.

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1.4 Determining and interpreting the nature of the rootsusing the discriminant

We already know from Topic 18: Identifying the features of a quadratic function how toidentify the shape of a quadratic graph but this does not tell us its position with respectto the x-axis. The discriminant will help us to determine this.



Key point

The discriminant of the quadratic equation ax2 + 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 respectto 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 onepoint.



Real and Distinct Roots

The discriminant is greater than zero.

The graph crosses the x-axis at twodistinct points.

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Here are some examples of both the graphs and the expressions in relation to thediscriminant.

Roots and the discriminant

Go onlineAn equation with no real roots has a discriminant with a negative value.

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.

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Examples

1. Real and distinct roots

Problem:

Determine the nature of the roots of the quadratic y = x2 − 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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2. Real and equal roots

Problem:

Determine the nature of the roots of the quadratic y = x2 + 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 thinkof this quadratic as having 1 solution.


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3. No real roots

Problem:

Find the roots of the quadratic y = x2 − 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 thesquare root of a negative number.


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Determining and interpreting the nature of the roots using thediscriminant exercise

Go onlineQ22:

Examine the discriminant of these quadratics, determine the nature of the roots andstate 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

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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

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The discriminant can tell us more about the roots.

Example

Problem:

Prove that the roots of the equation 5x2 + 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 x − 1 = 0

5x = −7 or x = 1

The solutions − 75 and 1 are both rational numbers.



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 realand rational. This is because the discriminant is the part of the quadratic formula underthe square root.

If we were to find the solutions using the quadratic formula we would get,

x =−2 ± √

144

2 × 5

x =−2 + 12

10or x =

−2 − 12

10again we find that both solutions are rational.

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Note: It is quicker to calculate the discriminant to determine whether the roots arerational 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 3x2 − 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.√40 is an irrational number) we know

that the roots are real and irrational.

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Key point

If the discriminant is positive but not a square number then the roots are real andirrational.

Rational and irrational roots practice

Go online

Q24: Prove that the roots of the equation 2x2 + 3x − 1 = 0 are real and irrational.

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Q25: Prove that the roots of the equation 3x2 − x − 2 = 0 are real and rational.

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Rational and irrational roots exercise

Go onlineQ26: Prove that the roots of the equation 6x2 + 5x + 1 = 0 are real and rational.

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Q27: Prove that the roots of the equation 3x2 + 6x − 4 = 0 are real and irrational.

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Q28: Are the roots of the equation 5x2 + 6x − 3 = 0 real and rational or real andirrational?

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Q29: Are the roots of the equation 2x2 − 13x + 3 = 0 real and rational or real andirrational?

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1.5 Learning points

Solving a quadratic equation graphically

• The zeros or roots of a quadratic are the point(s) where the graph crosses thex-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 = ax2 + bx + c must be factorised to takethe 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 = ax2 + bx + c you must identify thevalues of a, b and c.

• Substitute a, b and c into the quadratic formula.

• x = −b±√b2−4ac2a

• Remember to split the expression into:

◦ x = −b+√b2−4ac2a and

◦ x = −b−√b2−4ac2a

Identifying and interpreting the nature of the roots of a quadratic using thediscriminant

• From a quadratic equation of the form y = ax2 + bx + c you must identify thevalues 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.



◦ 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.



1.6 End of topic test

End of topic 19 test

Go online

Solving Quadratic Equations Graphically

Q30:

The diagram shows the graph of the function y = x2 + 3x − 4.Use the graph to solve the equation y = x2 + 3x − 4.

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Q31:

The diagram shows the graph of the function y = 12 + x − x2.Use the graph to solve the equation y = 12 + x − x2.

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Q32:

The diagram shows the graph of the function y = 2x2 − 11x + 9.Use the graph to solve the equation y = 2x2 − 11x + 9.



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Solving Quadratic Equations by Factorising

Q33: Solve x2 + 6x = 0

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Q34: Solve x2 − 25 = 0

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Q35: Solve x2 − 2x − 15 = 0

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Q36: Solve 3 − 2x − x2 = 0

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Q37: Solve 4x2 + 8x + 3 = 0

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Solving Quadratic Equations using the Quadratic Formula

Q38: Solve x2 + 7x + 8 = 0 using the quadratic formula, give your answer correct to1 d.p.

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Q41: Solve 2x2 − 2x − 4 = 0 using the quadratic formula, give your answer correctto 1 d.p.

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The Discriminant

Q42:


• 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

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Q43:


• real and distinct

• real and equal

• no real roots

a) 8x2 + 9x

b) 8x2 − 9

c) 8 − 8x + 2x2

d) 28 + 7x − x2

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Q44: Are the roots of the equation x2 − 7x + 3 = 0 are real and rational or real andirrational?

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22 GLOSSARY

Glossary

discriminant

the discriminant of the quadratic equation ax2 + bx + c = 0 is b2 − 4ac


ANSWERS: TOPIC 19 23

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) = 0x − 2 = 0 or x + 3 = 0x = 2 or x = − 3

Q9: (2x − 3)(x + 6) = 02x − 3 = 0 or x + 6 = 02x = 3 or x = − 6x = 1 · 5The 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 ashaving only one solution.

Solving quadratic equations by factorisation exercise (page 8)

Q11: (x − 6)(x + 4)

Q12:


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)


Q15:

Steps:

• Factorise 3 − 2x − x2. (x + 3)(−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

we see that a = 1 b = 2 c = − 7

x =−b ± √

b2 − 4ac

2a

⇒ x =−2 ± √

22 − 4 × 1 × (−7)

2 × 1

=−2 ± √

4 + 28

2

⇒ x =−2 +

√32

2or x =

−2 − √32

2⇒ x = 1 · 8 and x = −3 · 8 (to 1 d.p.)

Q18:

From 2x2 − 5x + 1

we see that a = 2 b = − 5 c = 1



x =−b ± √

b2 − 4ac

2a

⇒ x =− (−5) ±

√(−5)2 − 4 × 2 × 1

2 × 2

=5 ± √

25 − 8

4

⇒ x =5 +

√17

4or x =

5 − √17

4⇒ x = 2 · 3 and x = 0 · 2 (to 1 d.p.)

Solving quadratic equations using the quadratic formula exercise (page 10)

Q19:

Steps:

• To solve the equation x2 − x − 4 = 0 you must substitute a = 1, b = − 1 andc = − 4 into the quadratic formula.

• To evaluate the quadratic formula, you must start by calculating b2 − 4ac. 17

Answer: x = − 1 · 6 and x = 2 · 6Q20: x = − 1 · 1 and x = 7 · 1Q21: x = − 1 and x = 2

Determining and interpreting the nature of the roots using the discriminantexercise (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.



Rational and irrational roots practice (page 15)

Q24:

b2 − 4ac = 32 − 4 × 2 × (−1)

= 9 + 8

= 17



Q25:

b2 − 4ac = (−1)2 − 4 × 3 × (−2)

= 1 + 24

= 25

Since 25 is positive and a square number (i.e. 52 = 25) we know that the roots are realand 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. 12 = 1) we know that the roots are realand rational.

Q27:

b2 − 4ac = 62 − 4 × 3 × (−4)

= 36 + 48

= 84



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:



• 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)


Q34:

Steps:

• Factorise x2 − 25 = 0. (x − 5)(x + 5)


Q35:

Steps:

• Factorise x2 − 2x − 15. (x − 5)(x + 3)


Q36:

Steps:

• Factorise 3 − 2x − x2. (x + 3)(−x + 1)


Q37:

Steps:

• Factorise 4x2 + 8x + 3. (2x + 3)(2x + 1)

Answer: x = − 0 · 5 and x = − 1 · 5

Q38:

Hint:



• To solve the equation x2 + 7x + 8 = 0, you must substitute a = 1, b = 7 andc = 8 into the quadratic formula.

Steps:

• To evaluate the quadratic formula, begin by calculating b2 − 4ac. b2 − 4ac = 17

Answer: x = − 5 · 6 and x = − 1 · 4Q39:

Hint:

• To solve the equation x2 − 6x − 8 = 0, you must substitute a = 1, b = − 6and c = − 8 into the quadratic formula.

Steps:


Answer: x = − 1 · 1 and x = 7 · 1Q40:

Hint:

• To solve the equation x2 − 3x − 8 = 0, you must substitute a = 1, b = − 3and c = − 8 into the quadratic formula.

Steps:


Answer: x = − 1 · 7 and x = 4 · 7Q41:

Hint:

• To solve the equation 2x2 − 2x − 4 = 0, you must substitute a = 2, b = − 2and c = − 4 into the quadratic formula.

Steps:



Q42:

a) real and equal; b2 − 4ac = 0.


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.




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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SCHOLAR Study Guide National 5 Mathematics Course ... · SCHOLAR Study Guide National 5 Mathematics Course Materials Topic 19: Solving quadratic equations Authored by: Margaret Ferguson