© boardworks ltd 2005 1 of 48 a6 quadratic equations ks4 mathematics

© Boardworks Ltd 2005 1 of 48

A6 Quadratic equations

KS4 Mathematics


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AA6.1 Solving quadratic equations by factorization

Contents


A6.2 Completing the square

A6.3 Using the quadratic formula

A6.4 Equations involving algebraic fractions

A6.5 Problems leading to quadratic equations


Find the width of the rectangle

The length of this rectangle is 4 cm more than its width. The area of the rectangle is 45 cm2.

Find the width of the rectangle.

If we call the width of the rectangle x we can draw the following diagram:

x

x + 4

Using the information about the area of the rectangle we can write an equation:

x(x + 4) = 45


Find the width of the rectangle

The solution to x(x + 4) = 45 will give us the width of the rectangle.

In this example, it should be quite easy to spot that x = 5 is a possible solution to this equation, because

5 × 9 = 45

The width of the rectangle is therefore 5 cm.

However, there is another value of x that will also solve the equation x(x + 4) = 45.

This is because x(x + 4) = 45 is an example of a quadratic equation.


Quadratic equations

This is easier to spot if we multiply out the bracket,

x2 + 4x = 45

We usually arrange quadratic equations so that all the terms are on the left-hand side of the equals sign, leaving a 0 on the right-hand side. In this example we would have

The general form of a quadratic equation is

x(x + 4) = 45

x2 + 4x – 45 = 0

ax2 + bx + c = 0ax2 + bx + c = 0

Where a, b and c are constants and a ≠ 0.


Quadratic equations

We can solve the quadratic equation x2 + 4x – 45 = 0 in full by factorizing the expression on the left-hand side.

This means that we can write the equation in the form

(x + ….)(x + ….) = 0

We need to find two integers that add together to make 4 and multiply together to make –45.

Because –45 is negative, one of the numbers must be positive and one must be negative.

By considering the factors of 45 we find that the two numbers must be 9 and –5.

We can therefore write x2 + 4x – 45 = 0 as

(x + 9)(x – 5) = 0


Quadratic equations

When two numbers multiply together to make 0, one of the numbers must be 0, so if

we can conclude that either

This gives us two solutions that solve the quadratic equation:

(x + 9)(x – 5) = 0

x + 9 = 0 or x – 5 = 0

x = – 9 and x = 5

In the context of finding the width of a rectangle we cannot allow a negative length, and so x = 5 is the only valid solution.

Many other problems that lead to quadratic equations, however, would require both solutions.


Solving quadratic equations by factorization

Start by rearranging the equation so that the terms are on the left-hand side,

Factorizing the left-hand side gives us

Solve the equation x2 = 3x by factorization.

x2 – 3x = 0

x(x – 3) = 0

or x – 3 = 0

x = 3

x = 0So



Start by rearranging the equation so that the terms are on the left-hand side.

We need to find two integers that add together to make –5 and multiply together to make 4.

Factorizing the left-hand side gives us

Solve the equation x2 – 5x = –4 by factorization.

x2 – 5x + 4 = 0

(x – 1)(x – 4) = 0

x – 1 = 0 or x – 4 = 0

x = 4

Because 4 is positive and –5 is negative, both the integers must be negative. These are –1 and –4.

x = 1


Demonstrating solutions using graphs


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Contents




A6.1 Solving quadratic equations by factorization



Perfect squares

Some quadratic expressions can be written as perfect squares. For example,

x2 + 2x + 1 = (x + 1)2

x2 + 4x + 4 = (x + 2)2

x2 + 6x + 9 = (x + 3)2

x2 – 2x + 1 = (x – 1)2

x2 – 4x + 4 = (x – 2)2

x2 – 6x + 9 = (x – 3)2

How could the quadratic expression x2 + 6x be made into a perfect square?

We could add 9 to it.

In general,

x2 + 2ax + a2 = (x + a)2 and x2 – 2ax + a2 = (x – a)2


Completing the square

Adding 9 to the expression x2 + 6x to make it into a perfect square is called completing the square.

x2 + 6x = x2 + 6x + 9 – 9We can write

If we add 9 we then have to subtract 9 so that both sides are still equal.

By writing x2 + 6x + 9 we have completed the square and so we can write this as

x2 + 6x = (x + 3)2 – 9

x2 + bx = x + 2 – 2b2

b2

In general,



Complete the square for x2 – 10x.

Compare this expression to (x – 5)2 = x2 – 10x + 25

= (x – 5)2 – 25

x2 – 10x = x2 – 10x + 25 – 25

Complete the square for x2 – 3x.

Compare this expression to (x – 1.5)2 = x2 – 3x + 2.25

= (x + 1.5)2 – 2.25

x2 – 3x = x2 – 3x + 2.25 – 2.25


Expressions in the form x2 + bx



How can we complete the square for x2 + 8x + 9?

= (x + 4)2 – 7

x2 + 8x + 9 = x2 + 8x + 16 – 16 + 9

Look at the coefficient of x.

This is 8 so compare the expression to (x + 4)2 = x2 + 8x + 16.

x2 + bx + c = x + 2 – 2 + cb2

b2

In general,



Complete the square for x2 + 12x – 5.

Compare this expression to (x + 6)2 = x2 + 12x + 36

= (x2 + 6) – 41

x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5

Complete the square for x2 – 5x + 16

Compare this expression to (x – 2.5)2 = x2 – 5x + 6.25

= (x2 – 2.5) + 9.75

x2 – 5x + 16 = x2 – 5x + 6.25 – 6.25 + 16


Expressions in the form x2 + bx + c



When the coefficient of x2 is not 1, quadratic equations in the form ax2 + bx + c can be rewritten in the form a(x + p)2 + q by completing the square.

Complete the square for 2x2 + 8x + 3.

2x2 + 8x + 3 = 2(x2 + 4x) + 3

By completing the square, x2 + 4x = (x + 2)2 – 4 so,

2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3

= 2(x + 2)2 – 8 + 3

= 2(x + 2)2 – 5

Start by factorizing the first two terms by 2,



Complete the square for 5 + 6x – 3x2.

By completing the square, x2 – 2x = (x – 1)2 – 1 so,

5 + 6x – 3x2 = 5 – 3((x – 1)2 – 1)

= 5 – 3(x – 1)2 + 3

= 8 – 3(x – 1)2

Start by factorizing the the terms containing x’s by –3.

5 + 6x – 3x2 = 5 – 3(–2x + x2)

5 + 6x – 3x2 = 5 – 3(x2 – 2x)


Expressions in the form ax2 + bx + c


Solving quadratics by completing the square

Quadratic equations that cannot be solved by factorization can be solved by completing the square.

For example, the quadratic equation,

x2 – 4x – 3 = 0

can be solved by completing the square as follows,

(x – 2)2 – 4 – 3 = 0

x = 2 + √7 or x = 2 – √7

x = 4.646 x = –0.646 (to 3 d.p.)

x – 2 = ±√7square root both sides:

(x – 2)2 – 7 = 0simplify:

(x – 2)2 = 7add 7 to both sides:



Solve the equation x2 + 8x + 5 = 0 by completing the square. Write the answer to 3 decimal places.

x2 + 8x + 5 = 0

Completing the square on the left-hand side,

(x + 4)2 – 16 + 5 = 0

square root both sides: x + 4 = ±√11

x = –4 + √11 or x = –4 – √11

x = –0.683 x = –7.317 (to 3 d.p.)

(x + 4)2 = 11add 11 to both sides:

(x + 4)2 – 11 = 0simplify:



Solve the equation 2x2 – 4x + 1 = 0 by completing the square. Write the answer to 3 decimal places.

= 2((x – 1)2 – 1) + 1

= 2(x – 1)2 – 2 + 1

= 2(x – 1)2 – 1

We can complete the square for 2x2 – 4x + 1 by first factorizing the terms containing x’s by the coefficient of x2,

2x2 – 4x + 1 = 2(x2 – 2x) + 1

Next complete the square for the expression in the bracket,

We can now use this to solve the equation 2x2 – 4x + 1 = 0.



Solve the equation 2x2 – 4x + 1 = 0 by completing the square. Write the answer to 3 decimal places.

2x2 – 4x + 1 = 0

2(x – 1)2 = 1add 1 to both sides:

completing the square: 2(x – 1)2 – 1 = 0

divide both sides by 2: (x – 1)2 =12

square root both sides: x – 1 = ±12

x = 1 + 12

or x = 1 – 12

x = 1.707 x = 0.293 (to 3 d. p)


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Using the quadratic formula

Any quadratic equation of the form,

can be solved by substituting the values of a, b and c into the formula,

ax2 + bx + c = 0ax2 + bx + c = 0

x =–b ± b2 – 4ac

2a

This equation can be derived by completing the square on the general form of the quadratic equation.



Use the quadratic formula to solve x2 – 7x + 8 = 0.

1x2 – 7x + 8 = 0

x =–b ± b2 – 4ac

2a

x =2 × 1

7 ± (–7)2 – (4 × 1 × 8)

x =2

7 ± 49 – 32

x =2

7 + 17or x =

27 – 17

x = 5.562 x = 1.438 (to 3 d.p.)



Use the quadratic formula to solve 2x2 + 5x – 1 = 0.

2x2 + 5x – 1 = 0

x =–b ± b2 – 4ac

2a

x =2 × 2

–5 ± 52 – (4 × 2 × –1)

x =4

–5 ± 25 + 8

x =4

–5 + 33or x =

4–5 – 33

x = 0.186 x = –2.686 (to 3 d.p.)



Use the quadratic formula to solve 9x2 – 12x + 4 = 0.

9x2 – 12x + 4 = 0

x =–b ± b2 – 4ac

2a

x =2 × 9

12 ± (–12)2 – (4 × 9 × 4)

x =18

12 ± 144 – 144

x =18

12 ± 0

There is only one solution, x = 23



Use the quadratic formula to solve x2 + x + 3 = 0.

1x2 + 1x + 3 = 0

x =–b ± b2 – 4ac

2a

x =2 × 1

–1 ± 12 – (4 × 1 × 3)

x =2

–1 ± 1 – 12

x =2

–1 ± –11

We cannot find –11 and so there are no solutions.


Using b2 – 4ac

From using the quadratic formula,

x =–b ± b2 – 4ac

2a

we can see that we can use the expression under the square root sign, b2 – 4ac, to decide how many solutions there are.

When b2 – 4ac is negative, there are no solutions.

When b2 – 4ac is positive, there are two solutions.

When b2 – 4ac is equal to zero, there is one solution.


Using b2 – 4ac

We can demonstrate each of these possibilities using graphs.Remember, if we plot the graph of y = ax2 + bx + c the solutions to the equation ax2 + bx + c = 0 are given by the points where the graph crosses the x-axis.

y

x

b2 – 4ac is zero

One solution

y

x

b2 – 4ac is negative

No solutions

y

x

b2 – 4ac is positive

Two solutions


Using b2 – 4ac


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Equations involving algebraic fractions

Some equations involving algebraic fractions rearrange to give quadratic equations. For example:

The first step when solving equations involving fractions is to multiply through by the product of the denominators.

1x

+5

x + 4= 2

multiply by x(x + 4): x + 4 + 5x = 2x(x + 4)

collect all terms on the r.h.s.: 0 = 2x2 + 2x – 4

6x + 4 = 2x2 + 8xexpand brackets and simplify:

factorize: 0 = (x + 2)(x – 1)

x = –2 or x = 1

divide by 2: 0 = x2 + x – 2



Solve 4x + 2

–3

x + 8= 1

Start by multiplying through by x + 2 and x + 8 to remove the denominators,

4(x + 8) – 3(x + 2) = (x + 2)(x + 8)

expand the brackets: 4x + 32 – 3x – 6 = x2 + 10x + 16

collect all terms on the r.h.s.: 0 = x2 + 9x – 10

x + 26 = x2 + 10x + 16simplify:

factorize: 0 = (x + 10)(x – 1)

x = –10 or x = 1



Multiply through by x(4 – x): x2 + 2(4 – x) = 3x(4 – x)

collect terms on the l.h.s.: 4x2 – 14x + 8 = 0

x2 + 8 – 2x = 12x – 3x2expand the brackets:

divide by 2: 2x2 – 7x + 4 = 0

Solve,x

4 – x–

2x

= 3

This quadratic equation cannot be solved by factorization so we have to solve it using the quadratic formula.

Equations involving algebraic fractions may also lead to quadratic equations that do not factorize. For example,



Using the quadratic formula to solve 2x2 – 7x + 4 = 0

x =–b ± b2 – 4ac

2a

x =2 × 2

7 ± 72 – (4 × 2 × 4)

x =4

7 ± 49 – 32

x =4

7 + 17or x =

47 – 17

x = 2.781 x = 0.719 (to 3 d.p.)


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Problems leading to quadratic equations

Some real-life problems can be solved using quadratic equations.

If her total journey time to work and back is 1 hour, what was her average speed on the way to work?

Remember, time taken =distance

average speed

Let Jenny’s average speed on the way to work be x.

For example, Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour more slowly than on the way there.



Jenny drives 24 miles to get to work. On the way home she is caught in traffic and drives 20 miles per hour more slowly than on the way there. If her total journey time to work and back is 1 hour, what was her average speed on the way to work?

Jenny’s time taken to get to work =24x

Jenny’s time taken to get home from work =24

x – 20

Total time there and back =24

x – 2024x

+ = 1

Solving this equation will give us the value of x, Jenny’s average speed on the way to work.



24x – 20

24x

+ = 1

Start by multiplying through by x(x – 20) to remove the fractions:

24(x – 20) + 24x = x(x – 20)

expand the brackets: 24x – 480 + 24x = x2 – 20x

simplify: 48x – 480 = x2 – 20x

collect terms on the r.h.s.: 0 = x2 – 68x + 480

factorize: 0 = (x – 60)(x – 8)

We have two solutions x = 60 and x = 8.

Which of these solutions is not possible in this situation?



If Jenny’s average speed on the way to work was 8 miles per hour her average speed on the way home would be –12 miles per hour, a negative number.

The only solution that makes sense is x = 60 miles per hour.

This is usually because many physical quantities, such as length, can only be positive.

When practical problems lead to quadratic equations it is very often the case that only one of the solution will make sense in the context of the original problem.

We can therefore ignore the second solution.



The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the lengths of all three sides of the triangle.

Let’s start by drawing a diagram,

We can use Pythagoras’ Theorem to write an equation in terms of x.

x + 1x – 7

x



The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the length of all three sides of the triangle.

x2 + (x – 7)2 = (x + 1)2

x2 + (x – 7)(x – 7) = (x + 1)(x + 1)

expand: x2 + x2 – 7x – 7x + 49 = x2 + x + x + 1

simplify: 2x2 – 14x + 49 = x2 + 2x + 1

collect on the l.h.s.: x2 – 16x + 48 = 0

factorize: (x – 4)(x – 12) = 0

x = 4 or x = 12



The lengths of the two shorter sides in a right-angled triangle are x cm and (x – 7) cm. If the length of the hypotenuse is (x + 1) cm, find the value of x and hence the length of all three sides of the triangle.

If x = 4 then the lengths of the three sides are,

4 cm, 4 – 7 = –3 cm and 4 + 1 = 5 cm

We cannot have a side of negative length and so x = 4 is not a valid solution.

If x = 12 then the lengths of the three sides are,

12 cm, 12 – 7 = 5 cm and 12 + 1 = 13 cm

So, the shorter sides are 12 cm and 5 cm and the hypotenuse is 13 cm.

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