Quadratic s

Quadraticequations

13Chapter

Contents: A

B

C

D

E

F

Quadratic equations of the formx k2 =

The Null Factor law

Completing the square

Problem solving (Quadratic equations)

The quadratic formula (Extension)

More difficult equations (Extension)

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Y:\HAESE\IB_PD\IBPD_13\317ibpd13.cdr Monday, 20 March 2006 4:10:23 PM PETERDELL

OPENING PROBLEM

�

Equation ax2 + bx+ c = 0 form a b c Solutions

x2 ¡ 4 = 0 x2 + 0x¡ 4 = 0 1 0 ¡4 x = 2 or x = ¡2 two

(x¡ 2)2 = 0 x2 ¡ 4x+ 4 = 0 1 ¡4 4 x = 2 one

x2 + 4 = 0 x2 + 0x+ 4 = 0 1 0 4 none as x2 is always > 0 zero

Now consider the example, x2 + 3x¡ 10 = 0

if x = 2, x2 + 3x¡ 10

= 22 + 3£ 2 ¡ 10

= 4 + 6¡ 10

= 0

and if x = ¡5, x2 + 3x¡ 10

= (¡5)2 + 3£¡5 ¡ 10

= 25¡ 15¡ 10

= 0

Because x = 2 and x = ¡5 both satisfy the equation x2+3x¡10 = 0 we say that they

are solutions of it.

But, how do we find these solutions without using trial and error?

R = 12:5x2 ¡ 550x+ 8125 dollars.

How many jackets must be made and sold each week

in order to obtain income of $3000 each week?

Clearly we need to solve the equation:

12:5x2 ¡ 550x+ 8125 = 3000

i.e., 12:5x2 ¡ 550x+ 5125 = 0

How can we solve this equation?

Acme Leather Jacket Co. makes and sells leather jackets each day andtheir revenue function is given by

x

318 QUADRATIC EQUATIONS (Chapter 13)

Equations of the form ax+ b = 0 are called linear equations and have only one solution.

For example, 3x¡ 2 = 0 is a linear equation (a = 3 and b = ¡2) and has the one

solution, x = 2

3.

Quadratic equations may have two, one or zero solutions. We can demonstrate this with some

examples.

Quadratic equations can be written in the form ax2 + bx+ c = 0.


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Y:\HAESE\IB_PD\IBPD_13\318ibpd13.cdr Friday, 24 March 2006 2:31:48 PM PETERDELL

Consider the equation x2 ¡ 7 = 0

i.e., x2 = 7 (adding 7 to both sides)

Notice thatp

7£p7 = 7, so x =p

7 is one solution

and (¡p7)£ (¡p7) = 7, so x = ¡p7 is also a solution.

Thus, if x2 = 7, then x = §p7:

If x2 = k then

8><>:x = §pk if k > 0

x = 0 if k = 0

there are no real solutions if k < 0

1 Solve for x:

a x2 = 16

d 12x2 = 72

g 2x2 + 1 = 19

b 2x2 = 18

e 3x2 = ¡12

h 1¡ 3x2 = 10

c 3x2 = 27

f 4x2 = 0

i 2x2 + 7 = 13

§�~`7

7

is read as

‘plus or minus the

square root of ’

QUADRATIC EQUATIONS

OF THE FORM x k2=

A

SOLUTION OF (REVIEW)x k2=

Solve for x: a 2x2 + 1 = 15 b 2¡ 3x2 = 8

a 2x2 + 1 = 15

) 2x2 = 14 ftake 1 from both sidesg) x2 = 7 fdivide both sides by 2g) x = §p7

b 2¡ 3x2 = 8

) ¡3x2 = 6 ftake 2 from both sidesg) x2 = ¡2 fdividing both sides by ¡3g

which has no solutions as x2 cannot be < 0:

Example 1

EXERCISE 13A

QUADRATIC EQUATIONS (Chapter 13) 319


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Solve for x:

a (x¡ 3)2 = 16 b (x+ 2)2 = 11

a (x¡ 3)2 = 16

) x¡ 3 = §p16) x¡ 3 = §4

) x = 3§ 4

) x = 7 or ¡1

b (x+ 2)2 = 11

) x+ 2 = §p11

) x = ¡2§p11

2 Solve for x:

a (x¡ 2)2 = 9 b (x+ 4)2 = 25 c (x+ 3)2 = ¡1

d (x¡ 4)2 = 2 e (x+ 3)2 = ¡7 f (x+ 2)2 = 0

g (2x+ 5)2 = 0 h (3x¡ 2)2 = 4 i 1

3(2x¡ 1)2 = 8

Did you notice that for

equations of the form

we did not expand the

LHS?

( ) =x a k§ 2

Example 2


When the product of two (or more) numbers is zero, then at least one of them must be zero,

i.e., if ab = 0 then a = 0 or b = 0.

Solve for x using the Null Factor law:

a 3x(x¡ 5) = 0 b (x¡ 4)(3x+ 7) = 0

a 3x(x¡ 5) = 0

) 3x = 0 or x¡ 5 = 0

) x = 0 or 5

b (x¡ 4)(3x+ 7) = 0

) x¡ 4 = 0 or 3x+ 7 = 0

) x = 4 or 3x = ¡7

) x = 4 or ¡7

3

1 Solve for the unknown using the Null Factor law:

a 3x = 0

e ¡7y = 0

i x2 = 0

b 5y = 0

f ab = 0

j a2 = 0

c a£ 8 = 0

g 2xy = 0

k pqrs = 0

d b£¡2 = 0

h abc = 0

l a2b = 0

2 Solve for x using the Null Factor law:

a x(x+ 3) = 0 b 2x(x¡ 5) = 0 c (x¡ 1)(x¡ 3) = 0

d 4x(2¡ x) = 0 e ¡3x(2x+ 1) = 0 f 5(x+ 2)(2x¡ 1) = 0

g x2 = 0 h 2(x¡ 3)2 = 0 i ¡4(2x¡ 1)2 = 0

EXERCISE 13B

Example 3

THE NULL FACTOR LAWB


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To use the Null Factor law when solving equations, we must have one side of the equation

equal to zero.

Step 1: If necessary rearrange the equation with one side being zero.

Step 2: Fully factorise the other side (usually the LHS).

Step 3: Use the Null Factor law.

Step 4:

Step 5: Check at least one of your solutions.

Solve for x: x2 = 3x

x2 = 3x

) x2 ¡ 3x = 0 f‘equating to zero’ i.e., RHS= 0g) x(x¡ 3) = 0 ffactorising the LHSg

) x = 0 or x¡ 3 = 0 fNull Factor lawg) x = 0 or x = 3

) x = 0 or 3

Let us reconsider the equation x2 = 3x from Example 4.

If we cancel x from both sides we havex2

x=

3x

xand we finish with x = 3:

Consequently, we have ‘lost’ the solution x = 0.

From this example we conclude that:

3 Solve for x:

a x2 ¡ 7x = 0 b x2 ¡ 5x = 0 c x2 = 8x

d x2 = 4x e 3x2 + 6x = 0 f 2x2 + 5x = 0

g 4x2 ¡ 3x = 0 h 4x2 = 5x i 3x2 = 9x

Ifthen either

or .

a b

a b

£ = 0

= 0 = 0

STEPS FOR SOLVING QUADRATIC EQUATIONS

Example 4

ILLEGAL CANCELLING


Solve the resulting linear equations.

We must never cancel a variable that is a common factor from both sides of anequation unless we know that the factor cannot be zero.


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Solve for x: x2 + 3x = 28

x2 + 3x = 28

) x2 + 3x¡ 28 = 0 fone side must be 0g) (x+ 7)(x¡ 4) = 0 fas sum = +3 and product = ¡28

gives +7 and ¡4g) x+ 7 = 0 or x¡ 4 = 0 fNull Factor lawg

) x = ¡7 or 4 fsolving linear equationsg

4 Solve for x:

a x2 ¡ 1 = 0

d (x+ 2)2 = 0

g x2 + 5x+ 6 = 0

j x2 + 9x+ 14 = 0

m x2 + 4x = 12

5 Solve for x:

a x2 + 9x+ 14 = 0

d x2 + x = 12

g x2 = x+ 6

j 10¡ 3x = x2

b x2 ¡ 9 = 0

e x2 + 3x+ 2 = 0

h x2 ¡ 5x+ 6 = 0

k x2 + 11x = ¡30

n x2 = 11x¡ 24

b x2 + 11x+ 30 = 0

e x2 + 6 = 5x

h x2 = 7x+ 60

k x2 + 12 = 7x

c (x¡ 5)2 = 0

f x2 ¡ 3x+ 2 = 0

i x2 + 7x+ 6 = 0

l x2 + 2x = 15

o x2 = 14x¡ 49

c x2 + 2x = 15

f x2 + 4 = 4x

i x2 = 3x+ 70

l 9x+ 36 = x2

Solve for x: 5x2 = 3x+ 2

5x2 = 3x+ 2

) 5x2 ¡ 3x¡ 2 = 0 fmaking the RHS = 0g) 5x2 ¡ 5x+ 2x¡ 2 = 0 fac = ¡10, b = ¡3

) numbers are ¡5 and +2g) 5x(x¡ 1) + 2(x¡ 1) = 0

) (x¡ 1)(5x+ 2) = 0 ffactorisingg) x¡ 1 = 0 or 5x+ 2 = 0 fNull Factor lawg

) x = 1 or ¡2

5fsolving the linear equationsg

6 Solve for x:

a 2x2 + 2 = 5x

d 2x2 + 5x = 3

g 3x2 + 13x+ 4 = 0

j 2x2 + 3x = 5

b 3x2 + 8x = 3

e 2x2 + 5 = 11x

h 5x2 = 13x+ 6

k 3x2 + 2x = 8

c 3x2 + 17x+ 20 = 0

f 2x2 + 7x+ 5 = 0

i 2x2 + 17x = 9

l 2x2 + 9x = 18

Example 5

Example 6



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Y:\HAESE\IB_PD\IBPD_13\322ibpd13.cdr Friday, 24 February 2006 10:44:04 AM PETERDELL

Solve for x: 10x2 ¡ 13x¡ 3 = 0

10x2 ¡ 13x¡ 3 = 0 fac = ¡30, b = ¡13) numbers are ¡15 and 2g) 10x2 ¡ 15x+ 2x¡ 3 = 0

) 5x(2x¡ 3) + 1(2x¡ 3) = 0 ffactorising in pairsg) (2x¡ 3)(5x+ 1) = 0 fcommon factor factorisationg

) 2x¡ 3 = 0 or 5x+ 1 = 0 fusing the Null Factor lawg) x = 3

2or ¡1

5fsolving the linear equationsg

7 Solve for x:

a 6x2 + 13x = 5

d 21x2 = 62x+ 3

b 6x2 = x+ 2

e 10x2 + x = 2

c 6x2 + 5x+ 1 = 0

f 10x2 = 7x+ 3

8 Solve for x by first expanding brackets and then equating to zero:

a x(x+ 5) + 2(x+ 6) = 0 b x(1 + x) + x = 3

c (x¡ 1)(x+ 9) = 8x d 3x(x+ 2)¡ 5(x¡ 3) = 17

e 4x(x+ 1) = ¡1 f 2x(x¡ 6) = x¡ 20

Example 7

COMPLETING THE SQUAREC


Try as much as we like, we will not be able to solve quadratic equations such as

x2 + 4x ¡ 7 = 0 by using the factorisation methods already practised. This is because the

solutions are not rationals.

To solve this equation we need a different technique.

Consider the solution to the equation (x+ 2)2 = 11

) x+ 2 = §p11

) x = ¡2§p11

However, if (x+ 2)2 = 11

then x2 + 4x+ 4 = 11

and x2 + 4x¡ 7 = 0

and so the solutions to x2 + 4x¡ 7 = 0 are the solutions to (x+ 2)2 = 11:

Consequently, an approach for solving such equations could be to reverse the above argument.

Consider x2 + 4x¡ 7 = 0

) x2 + 4x = 7

) x2 + 4x+ 4 = 7 + 4

) (x+ 2)2 = 11

) x+ 2 = §p11

) x = ¡2§p11

Hence the solutions to x2 + 4x¡ 7 = 0 are x = ¡2§p11.


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From the above example it can be seen that a perfect square needs to be created on the left

hand side. The process used is called completing the square.

From our previous study of perfect squares we observe that:

(x+ 3)2 = x2 + 2£ 3£ x+ 32 notice that 32 =

µ2£ 3

2

¶2

(x¡ 5)2 = x2 ¡ 2£ 5£ x+ 52 notice that 52 =

µ2£ 5

2

¶2

(x+ p)2 = x2 + 2£ p£ x+ p2 notice that p2 =

µ2£ p

2

¶2

i.e., the constant term is “the square of half the coefficient of x”.

To create a perfect square on the LHS, what must be added to both sides of

the equation a x2 + 8x = ¡5 b x2 ¡ 6x = 13?

What does the equation become in each case?

a In x2 + 8x = ¡5, half the coefficient of x is 8

2= 4

so, we add 42 to both sides

and the equation becomes x2 + 8x+ 42 = ¡5 + 42

(x+ 4)2 = ¡5 + 16

(x+ 4)2 = 11

b In x2 ¡ 6x = 13, half the coefficient of x is ¡6

2= ¡3

so, we add (¡3)2 = 32 to both sides

and the equation becomes x2 ¡ 6x+ 32 = 13 + 32

(x¡ 3)2 = 13 + 9

(x¡ 3)2 = 22

1 For each of the following equations:

i

ii write each equation in the form (x+ p)2 = k

a x2 + 2x = 5

d x2 ¡ 6x = ¡3

g x2 + 12x = 13

b x2 ¡ 2x = ¡7

e x2 + 10x = 1

h x2 + 5x = ¡2

c x2 + 6x = 2

f x2 ¡ 8x = 5

i x2 ¡ 7x = 4

Notice that we keep theequation balanced by

adding the same to bothsides of the equation.

Example 8

find what must be added to both sides of the equation to create a perfectsquare on the LHS

EXERCISE 13C



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Solve for x by completing the square, leaving answers in surd form:

a x2 + 2x¡ 2 = 0 b x2 ¡ 4x+ 6 = 0

a x2 + 2x¡ 2 = 0

) x2 + 2x = 2 fmove constant term to RHSg) x2 + 2x+ 12 = 2 + 12 fadd (2

2)2 = 12 to both sidesg

) (x+ 1)2 = 3 ffactorise LHS, simplify RHSg) x+ 1 = §p3

) x = ¡1§p3

So, solutions are x = ¡1 +p

3 or ¡1¡p3:

b x2 ¡ 4x+ 6 = 0

) x2 ¡ 4x = ¡6 fremove the constant term to the RHSg) x2 ¡ 4x+ 22 = ¡6 + 22 fadd (¡4

2)2 = 22 to both sidesg

) (x¡ 2)2 = ¡2 ffactorise the LHS, simplify the RHSgwhich is impossible as no perfect square can be negative

) no real solutions exist.

2 If possible, solve for x using ‘completing the square’, leaving answers in surd form:

a x2 ¡ 4x+ 1 = 0

d x2 + 2x¡ 1 = 0

g x2 + 6x+ 3 = 0

3 Using the method of ‘completing the square’ solve for x, leaving answers in surd form.

a x2 + 3x+ 2 = 0

d x2 + x¡ 1 = 0

b x2 ¡ 2x¡ 2 = 0

e x2 + 2x+ 4 = 0

h x2 ¡ 6x+ 11 = 0

b x2 = 4x+ 8

e x2 + 3x¡ 1 = 0

c x2 ¡ 4x¡ 3 = 0

f x2 + 4x+ 1 = 0

i x2 + 8x+ 14 = 0

c x2 ¡ 5x+ 6 = 0

f x2 + 5x¡ 2 = 0

4 a To solve 3x2 + 6x¡ 1 = 0 by ‘completing the square’ our first step must be to

divide both sides by 3, and this results in x2 + 2x ¡ 1

3= 0. Explain why this

statement is true.

b Use ‘completing the square’ to solve

i 2x2 + 4x¡ 1 = 0 ii 3x2 ¡ 12x+ 7 = 0 iii 5x2 ¡ 10x+ 3 = 0

Remember that if

, where

then .

x k k >

x k

2= 0= §�~`

Example 9



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possible

original

numbers.

by squares

1 The sum of a number and its square is 90. Find the number.

2 The product of a number and the number decreased by 3 is 108. Find the two

answers for the number.

3 When 32 is subtracted from the square of a number the result is six times the

number. Find the number.

4 The sum of two numbers is 9 and the sum of their squares is 153. Find the

5 Two numbers differ 5 and the sum of their is 17. Find the numbers.

PROBLEM SOLVING

(QUADRATIC EQUATIONS)D

The sum of a number and its square is 42. Find the number.

Let the number be x. Therefore its square is x2.

x+ x2 = 42

) x2 + x¡ 42 = 0 frearrangingg) (x+ 7)(x¡ 6) = 0 ffactorisingg) x = ¡7 or x = 6

Check: If x = ¡7, ¡7 + (¡7)2 = ¡7 + 49 = 42 X

If x = 6, 6 + 62 = 6 + 36 = 42 X

So, the number is ¡7 or 6.

Example 10

EXERCISE 13D


Contained in this section are problems which when converted to algebraic form result in a

quadratic equation.

Step 1: Carefully read the question until you understand the problem.

A rough sketch may be useful.

Step 2: Decide on the unknown quantity, calling it x, say.

Step 3: Find an equation which connects x and the information you are given.

Step 4: Solve the equation using one of the methods you have learnt.

Step 5: Check that any solutions satisfy the equation and are realistic to the problem.

Step 6: Write your answer to the question in sentence form.

PROBLEM SOLVING METHOD

It is essential that you are successful at solving these equations using or‘completing the square’.

factorisation


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A rectangle has length 5 cm greater than its width. If it has an area of 84 cm2 ,

find the dimensions of the rectangle.

If x cm is the width, then (x+ 5) cm is the length.

Now area = 84 cm2

) x(x+ 5) = 84

) x2 + 5x = 84

) x2 + 5x¡ 84 = 0

) (x+ 12)(x¡ 7) = 0 fon factorisationg) x = ¡12 or 7:

But x > 0 as lengths are positive quantities, ) x = 7

) the rectangle is 7 cm by 12 cm.

6 A rectangle has length 5 cm greater than its width. Find its width given that its area is

150 cm2.

7 A triangle has base 2 cm more than its altitude. If its area is 49:5 cm2, find its altitude.

8 A rectangular enclosure is made from 50 m of fencing. The area enclosed is 144 m2.

Find the dimensions of the enclosure.

9 A rectangular pig pen is built against an existing brick

fence. 24 m of fencing was used to enclose 70 m2.

Find the dimensions of the pen.

10 Use the theorem of Pythagoras to find x given:

a b

11 A right angled triangle has sides 1 cm and 18 cm respectively less than its hypotenuse.

Find the length of each side of the triangle.

12 A forestry worker plants 600 pine trees. The number of pine trees in each row is 10more than twice the number of rows. If equal numbers of pine trees were planted in

each row, how many rows did the forestry worker plant?

13 ABCD is a rectangle in which AB = 21 cm.

The square AXYD is removed and the remaining

rectangle has area 80 cm2.

Find the length of BC.

Example 11

( 5) cmx��

x cm

D

A X

Y C

B

( 8) cmx�

( 7) cmx�

x cm

(2 2) cmx�

( 2) cmx�

x cm



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14 A, B, C and D are posts on the banks of a

20 m wide canal. A and B are 1 m apart.

If OA is the same length as CD, find how

far C and D are apart.

15 AB is 2 cm longer than BE. DC is 3 cm less

than twice BE.

a Explain why triangles ABE and ACD are

similar.

b If BE = x cm, show that x2 ¡ 4x¡ 6 = 0:

c Hence, show that BE = 2 +p

10 cm.

The sum of a number and four times its reciprocal is 812: Find the number.

Let the number be x. Therefore, its reciprocal is1

x:

x + 4£ 1

x= 81

2fas the sum is 81

2g

)x

1+

4

x=

17

2which has an LCD of 2x

)x

1£ 2x

2x+

4

x£ 2

2=

17

2£ xx

fto achieve a common denominatorg

) 2x2 + 8 = 17x fequating the numeratorsg) 2x2 ¡ 17x+ 8 = 0 fequating to zerog2x2 ¡ x¡ 16x+ 8 = 0 fsplitting the middle termg

x(2x¡ 1)¡ 8(2x¡ 1) = 0 ffactorising in pairsg) (2x¡ 1)(x¡ 8) = 0 fcommon factorg

2x¡ 1 = 0 or x¡ 8 = 0 fNull Factor lawgx = 1

2or 8

So, the number is 1

2or 8. Check: 1

2+ 4£ 2 = 81

2X 8 + 4£ 1

8= 81

2X

16 The sum of a number and its reciprocal is 2 1

12: Find the number.

17 The sum of a number and twice its reciprocal is 323

. Find the number.

A B C

D

E

3 cm

DC

B A

O

20 m

Example 12



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HISTORICAL NOTE BABYLONIAN ALGEBRA

18 A rectangular sheet of tin plate is 20 cm by

16 cm and is to be made into an open box

with a base having an area of 140 cm2, by

cutting out equal squares from the four cor-

ners and then bending the edges upwards.

Find the size of the squares cut out.

19 A rectangular swimming pool is 12 m long by 6 m wide. It is surrounded by a pavement

of uniform width, the area of the pavement being 7

8of the area of the pool.

a If the pavement is x m wide, show that the area of the pavement is 4x2+36x m2.

b Hence, show that 4x2 + 36x¡ 63 = 0.

c How wide is the pavement?

20 A circular magnet has an inner radius x cm, an

outer radius 2 cm larger and its depth is the

same as the inner radius (as shown).

If the total volume of the magnet is 120¼ cm3,

find x.

x cm

2 cm

x cm

The mathematics used by thewas recorded on clay tablets in cuneiform.One such tablet which has been preserved iscalled (around BC).

Babylonians

Plimpton 322 1600


They could, for example, add 4xy to (x¡y)2 to obtain (x+y)2. This was all achieved

without the use of letters for unknown quantities. However, they often used words for the

unknown.

Consider the following example from about 4000 years ago.

Problem: “I have subtracted the side of my square from the area and the result is 870.

What is the side of the square?”

Solution: Take half of 1, which is 1

2, and multiply 1

2by 1

2which is 1

4;

add this to 870 to get 87014

. This is the square of 2912

.

Now add 1

2to 291

2and the result is 30, the side of the square.

Using our modern symbols: the equation is x2 ¡ x = 870 and the solution is

x =q

(12)2 + 870 + 1

2= 30.

This solution is one of the two solutions we would obtain using the Quadratic Formula.

We meet this formula in the next section.

The Ancient Babylonians were able to solve difficultequations using the rules we use today, such as transposingterms and multiplying both sides by like quantities toremove fractions. They were familiar with factorisation.


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Y:\HAESE\IB_PD\IBPD_13\329ibpd13.cdr Tuesday, 21 March 2006 9:17:34 AM PETERDELL

INVESTIGATION THE QUADRATIC FORMULA (Extension)

1

a x2 + 3x = 10

d 4x2 ¡ 12x+ 9 = 0

b 2x2 = 7x+ 4

e x2 + 6x+ 11 = 0

c x2 = 4x+ 3

f 4x2 + 1 = 8x

2 A formula for solving ax2 + bx+ c = 0 is x =¡b§pb2 ¡ 4ac

2a:

Check that this formula gives the correct answer for each question in 1.

3 What is the significance of b2 ¡ 4ac in determining the solutions of the quadratic

equation ax2 + bx+ c = 0? You should be able to answer this question from obser-

vations in 1 and 2.

4 Establish the quadratic formula by ‘completing the square’ on ax2 + bx+ c = 0:

(Hint: Do not forget to divide each term by a to start with.)

Many quadratic equations cannot be solved by factorising, and completing the square is rather

tedious. Consequently, the quadratic formula has been developed. This formula is:

If ax2 + bx+ c = 0, then x =

¡b§pb2 ¡ 4ac

2a.

Consider the Opening problem involving the Acme Leather Jacket Co. The equation we

need to solve is:

12:5x2 ¡ 550x+ 5125 = 0 where a = 12:5, b = ¡550, c = 5125

Using the formula we obtain x =550§p46 250

25which simplifies to x + 30:60 or 13:40 .

But as x needs to be a whole number, x = 13 or 31 would produce income of around $3000each week.

The following proof of the quadratic formula is worth careful examination.

Proof: If ax2 + bx+ c = 0,

then x2 +b

ax+

c

a= 0 fdividing each term by a, as a 6= 0g

) x2 +b

ax = ¡ c

a

) x2 +b

ax+

µb

2a

¶2= ¡ c

a+

µb

2a

¶2fcompleting the square on LHSg

What to do:

Use factorisation techniques or ‘completing the square’ where necessaryto solve:

THE QUADRATIC FORMULA (EXTENSION)E



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)

µx+

b

2a

¶2= ¡ c

a

µ4a

4a

¶+b2

4a2

)

µx+

b

2a

¶2=b2 ¡ 4ac

4a2

) x+b

2a= §

rb2 ¡ 4ac

4a2

) x = ¡ b

2a§rb2 ¡ 4ac

4a2

i.e., x =¡b§pb2 ¡ 4ac

2a

Solve for x: a x2 ¡ 2x¡ 2 = 0 b 2x2 + 3x¡ 4 = 0

a x2 ¡ 2x¡ 2 = 0 has a = 1, b = ¡2, c = ¡2

) x =¡(¡2)§p(¡2)2 ¡ 4(1)(¡2)

2(1)

) x =2§p4 + 8

2

) x =2§p12

2

) x =2§ 2

p3

2

) x = 1§p3

So, the solutions are 1 +p

3 and 1¡p3:

b 2x2 + 3x¡ 4 = 0 has a = 2, b = 3, c = ¡4

) x =¡3§p32 ¡ 4(2)(¡4)

2(2)

) x =¡3§p9 + 32

4

) x =¡3§p41

4

So, the solutions are¡3 +

p41

4and

¡3¡p41

4:

Example 13



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1 Use the quadratic formula to solve for x:

a x2 ¡ 4x¡ 3 = 0

d x2 + 4x = 1

g x2 ¡ 2p

2x+ 2 = 0

2 Use the quadratic formula to solve for x:

a (x+2)(x¡1) = 2¡3x

b x2 + 6x+ 7 = 0

e x2 ¡ 4x+ 2 = 0

h (3x+ 1)2 = ¡2x

b (2x+ 1)2 = 3¡ x

c x2 + 1 = 4x

f 2x2 ¡ 2x¡ 3 = 0

i (x+ 3)(2x+ 1) = 9

c (x¡ 2)2 = 1 + x

The solutions are: x + ¡0:8508 or 2:351

Click on the appropriate icon for helpful instructions if

using a graphics calculator and/or graphing package.

1 Use technology to solve:

a x2 + 4x+ 2 = 0

d 3x2 ¡ 7x¡ 11 = 0

b x2 + 6x¡ 2 = 0

e 4x2 ¡ 11x¡ 13 = 0

c 2x2 ¡ 3x¡ 7 = 0

f 5x2 + 6x¡ 17 = 0

EXERCISE 13E.1

SOLVING USING TECHNOLOGY

EXERCISE 13E.2

y

x

y x x= 2 3 4X� �

321

-1

2

-2

-4

GRAPHING

PACKAGETI

C


A graphics calculator or graphing package can be used to solve quadratic equations. How-

ever, exact solutions in square root form will be lost in most cases. Approximate decimal

solutions are usually generated.

In this course we will find solutions using graphs of quadratics. We will examine their inter-

section with the x-axis, giving us zeros, and we will also examine where functions intersect,

finding the x-coordinates of the points where they meet.

We have chosen to use this approach, even though it may not be the quickest, so that an

understanding of the link between the algebra and the graphics is fully appreciated.

Consider the equation 2x2 ¡ 3x¡ 4 = 0.

Our approach will be:

² draw the graph of y = 2x2 ¡ 3x¡ 4

² 2x2 ¡ 3x¡ 4 = 0 when y = 0and this occurs at the -intercepts of

the graph.

x


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Y:\HAESE\IB_PD\IBPD_13\332ibpd13.cdr 24 March 2006 15:27:51 DAVID2

To solve a more complicated equation like (x¡ 2)(x+ 1) = 2 + 3x we could:

² make the RHS zero i.e., (x¡ 2)(x+ 1)¡ 2¡ 3x = 0:Plot y = (x¡ 2)(x+ 1)¡ 2¡ 3x and find the x-intercepts.

² Plot y = (x¡ 2)(x+ 1) and y = 2 + 3x on the same axes and

find the x-coordinates where the two graphs meet.

Y1 = (x¡ 2)(x+ 1) and Y2 = 2 + 3x we get

So, the solutions are x + ¡0:8284 or 4:8284

2 Use technology to solve:

a (x+2)(x¡1) = 2¡3x b (2x+ 1)2 = 3¡ x c (x¡ 2)2 = 1 + x

Consider x2 + 2x+ 5 = 0.

Using the quadratic formula, the solutions are: x =¡2§p4¡ 4(1)(5)

2(1)

i.e., x =¡2§p¡16

2

In the real number systemp¡16 does not exist and so we would say that x2+2x+5 has

no real solutions.

If we graph y = x2 + 2x+ 5 we get:

Clearly, the graph does not cut the x-axis and this

further justifies the fact that x2 + 2x + 5 = 0has no real solutions.

3 Solve for x:

a x2 ¡ 25 = 0

d x2 + 7 = 0

g x2 ¡ 4x+ 5 = 0

j x2 + 6x+ 25 = 0

b x2 + 25 = 0

e 4x2 ¡ 9 = 0

h x2 ¡ 4x¡ 5 = 0

k 2x2 ¡ 6x+ 5 = 0

c x2 ¡ 7 = 0

f 4x2 + 9 = 0

i x2 ¡ 10x+ 29 = 0

l 2x2 + x+ 1 = 0

QUADRATIC EQUATIONS WITH NO REAL SOLUTIONS

TI

C

GRAPHING

PACKAGE

y

x(-1' 4)

42-2-4-6

15

10

5


Using a graphics calculator with ,


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In the quadratic formula, b2 ¡ 4ac, which is under the square root sign, is called the

discriminant.

The symbol delta ¢, is used to represent the discriminant, i.e., ¢ = b2 ¡ 4ac.

The quadratic formula becomes x =¡b§p¢

2aif ¢ replaces b2 ¡ 4ac:

Notice that

4 What is the discriminant of:

a x2 ¡ 2x¡ 7 = 0

d 2x2 ¡ 6x¡ 4 = 0

b 2x2 ¡ 3x+ 6 = 0

e 3x2 + 7x¡ 1 = 0

c x2 ¡ 11 = 0

f 4x2 ¡ 7x+ 11 = 0

5 By using the discriminant only, state the nature of the solutions of:

a x2 + 7x¡ 2 = 0 b x2 + 4p

2x+ 8 = 0

d 6x2 + 5x¡ 4 = 0 e x2 + x+ 6 = 0

6 By using the discriminant only, determine which of the following quadratic equations

have rational roots which can be found by factorisation.

a 2x2 + 7x¡ 4 = 0 b 3x2 ¡ 7x¡ 6 = 0

d 6x2 + 19x+ 10 = 0 e 4x2 ¡ 3x+ 3 = 0

c 2x2 + 3x¡ 1 = 0

f 9x2 + 6x+ 1 = 0

c 2x2 + 6x+ 1 = 0

f 8x2 ¡ 10x¡ 3 = 0

THE DISCRIMINANT, ¢

Use the discriminant to determine the nature of the roots of:

a 2x2 ¡ 3x+ 4 = 0 b 4x2 ¡ 4x¡ 1 = 0

a b ¢ = b2 ¡ 4ac

= (¡4)2 ¡ 4(4)(¡1)

= 32 which is > 0

) has 2 distinct real roots

Example 14


² if ¢ = 0, x =¡2a

is the only solution (a repeated root)

² if ¢ > 0,p

¢ is a real number and so there are two distinct real roots,

¡b+p

¢

2aand

¡b¡p¢

2a

² if ¢ < 0,p

¢ does not exist so we have no real solutions.

Note: If , and are rational and then the equationhas two rational roots which can be found by factorisation.a b c ¢ is a perfect square

b

¢ = b2 ¡ 4ac

= (¡3)2 ¡ 4(2)(4)

= ¡23 which is < 0

) there are no real roots


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REVIEW SET 13A

Sometimes we need to simplify an equation before we can attempt to solve it.

In the following exercise we have rational equations.

1 Solve for x by first eliminating the algebraic fractions:

ax

3=

2

x

dx¡ 1

4=

3

x

g2x

3x+ 1=

1

x+ 2

jx¡ 1

2¡ x = 2x+ 1

b4

x=x

2

ex¡ 1

x=x+ 11

5

h2x+ 1

x= 3x

k x¡ 1

x= 1

cx

5=

2

x

fx

x+ 2=

1

x

ix+ 2

x¡ 1=x

2

l 2x¡ 1

x= 3

1 Solve for x:

a 2x2 = 4 b 3x2 + 18 = 0 c 5x(x¡ 3) = 0

d x2 + 24 = 11x e 10x2 ¡ 11x¡ 6 = 0 f 3x2 = 2x+ 21

2 Solve by ‘completing the square’: a x2 + 6x+ 11 = 0 b x2 ¡ 14x+ 7 = 0

3

4 Expand and simplify:

a 5¡ 2x¡ (x+ 3)2 b (x¡ 2)3 c (3x¡ 2)(x2 + 2x+ 7)

5 The sum of a number and its reciprocal is 216

. Find the number.

6

7 Solve for x: x2 ¡ 5x+ 2 = 0

8

a x2 + 3x¡ 6 = 0 b 2x2 + 5x+ 7 = 0

MORE DIFFICULT EQUATIONS(EXTENSION)

F

EXERCISE 13F

The width of a rectangle is cm less than its length and its area is cm . Find itsdimensions.

7 260 2

When the square of a number is increased by one the result is four times the originalnumber. Find the number.

Using the discriminant only, state the nature (how many, if they exist) of the solutions of:


We need to get rid of the algebraic fractions and then convert to a quadratic equation in eachcase.


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REVIEW SET 13B

1 Solve for x:

a ¡2(x¡ 3)2 = 0 b (x+ 5)(x¡ 4) = 0 c (2¡ x)2 = ¡1

d x2 + 5x = 24 e 2x2 ¡ 18 = 0 f 8x2 + 2x¡ 3 = 0

2 Solve by ‘completing the square’: a x2 ¡ 2x = 100 b x2 + x¡ 9 = 0

3 A rectangle has its length 3 cm greater than its width. If it has an area of 108 cm2,

find the dimensions of the rectangle.

4

5

6 Find b and c if x2 + bx+ c = 0 has solutions x = ¡1 or x = 7.

7 Solve for x: a x2 ¡ 2x+ 7 = 0 b 4x2 ¡ x+ 1 = 0

8

a x2 +p

3x+ 1 = 0 b 2x2 + 5x+ 2 = 0

When the square of a number is increased by , the result is seven times the originalnumber. Find the number.

10

A right angled triangle has its hypotenuse one centimetre more than twice the lengthof the shortest side, while the other side is cm longer than the shortest side. Find thelength of each of the sides of the triangle.

7

Using the discriminant only, state the nature (how many, if they exist) of the solutionsof:



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Documents

Quadratic s