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KS4 Mathematics
A6 Quadratic equations
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Contents
A6 Quadratic equations
A6.1 Solving quadratic equations by factorization
A6.2 Completing the square
A6.3 Using the quadratic formula
A6.4 Equations involving algebraic fractions
A6.5 Problems leading to quadratic equations
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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+4
x
Using the information about the area of the rectangle we can
write an equation:
x(x + 4) = 45
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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.
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Quadratic equations
This is easier to spot if we multiply out the bracket,
x(x + 4) = 45
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
x2 + 4x – 45 = 0
The general form of a quadratic equation is
ax2 + bx + c = 0
Where a, b and c are constants and a ≠ 0.
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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
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Quadratic equations
When two numbers multiply together to make 0, one of the
numbers must be 0, so if
(x + 9)(x – 5) = 0
we can conclude that either
x+9=0
or
x–5=0
This gives us two solutions that solve the quadratic equation:
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.
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Solving quadratic equations by factorization
Solve the equation x2 = 3x by factorization.
Start by rearranging the equation so that the terms are on the
left-hand side,
x2 – 3x = 0
Factorizing the left-hand side gives us
x(x – 3) = 0
So
x=0
or
x–3=0
x=3
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Solving quadratic equations by factorization
Solve the equation x2 – 5x = –4 by factorization.
Start by rearranging the equation so that the terms are on the
left-hand side.
x2 – 5x + 4 = 0
We need to find two integers that add together to make –5
and multiply together to make 4.
Because 4 is positive and –5 is negative, both the integers
must be negative. These are –1 and –4.
Factorizing the left-hand side gives us
(x – 1)(x – 4) = 0
x – 1 = 0 or
x–4=0
x=1
x=4
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Solving quadratic equations by factorization
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Demonstrating solutions using graphs
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Contents
A6 Quadratic equations
A6.1 Solving quadratic equations by factorization
A6.2 Completing the square
A6.3 Using the quadratic formula
A6.4 Equations involving algebraic fractions
A6.5 Problems leading to quadratic equations
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Perfect squares
Some quadratic expressions can be written as perfect squares.
For example,
x2 + 2x + 1 = (x + 1)2
x2 – 2x + 1 = (x – 1)2
x2 + 4x + 4 = (x + 2)2
x2 – 4x + 4 = (x – 2)2
x2 + 6x + 9 = (x + 3)2
x2 – 6x + 9 = (x – 3)2
In general,
x2 + 2ax + a2 = (x + a)2 and x2 – 2ax + a2 = (x – a)2
How could the quadratic expression x2 + 6x
be made into a perfect square?
We could add 9 to it.
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Completing the square
Adding 9 to the expression x2 + 6x to make it into a perfect
square is called completing the square.
We can write
x2 + 6x = x2 + 6x + 9 – 9
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
In general,
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x2
+ bx = x +
b 2
2
–
b 2
2
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Completing the square
Complete the square for x2 – 10x.
Compare this expression to (x – 5)2 = x2 – 10x + 25
x2 – 10x = x2 – 10x + 25 – 25
= (x – 5)2 – 25
Complete the square for x2 – 3x.
Compare this expression to (x – 1.5)2 = x2 – 3x + 2.25
x2 – 3x = x2 – 3x + 2.25 – 2.25
= (x + 1.5)2 – 2.25
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Expressions in the form x2 + bx
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Completing the square
How can we complete the square for
x2 + 8x + 9?
Look at the coefficient of x.
This is 8 so compare the expression to (x + 4)2 = x2 + 8x + 16.
x2 + 8x + 9 = x2 + 8x + 16 – 16 + 9
= (x + 4)2 – 7
In general,
x2
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+ bx + c = x +
b 2
2
–
b
2
2
+c
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Completing the square
Complete the square for x2 + 12x – 5.
Compare this expression to (x + 6)2 = x2 + 12x + 36
x2 + 12x – 5 = x2 + 12x + 36 – 36 – 5
= (x2 + 6) – 41
Complete the square for x2 – 5x + 16
Compare this expression to (x – 2.5)2 = x2 – 5x + 6.25
x2 – 5x + 16 = x2 – 5x + 6.25 – 6.25 + 16
= (x2 – 2.5) + 9.75
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Expressions in the form x2 + bx + c
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Completing the square
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.
Start by factorizing the first two terms by 2,
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
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Completing the square
Complete the square for 5 + 6x – 3x2.
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)
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
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Expressions in the form ax2 + bx + c
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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)2 – 7 = 0
simplify,
(x – 2)2 = 7
add 7 to both sides
square root both sides,
x = 2 + √7
x = 4.646
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x – 2 = ±√7
or
x = 2 – √7
x = –0.646 (to 3 d.p.)
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Solving quadratics by completing the square
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
(x + 4)2 – 11 = 0
simplify,
add 11 to both sides
(x + 4)2 = 11
square root both sides,
x + 4 = ±√11
x = –4 + √11
x = –0.683
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or
x = –4 – √11
x = –7.317 (to 3 d.p.)
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Solving quadratics by completing the square
Solve the equation 2x2 – 4x + 1 = 0 by completing
the square. Write the answer to 3 decimal places.
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,
= 2((x – 1)2 – 1) + 1
= 2(x – 1)2 – 2 + 1
= 2(x – 1)2 – 1
We can now use this to solve the equation 2x2 – 4x + 1 = 0.
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Solving quadratics by completing the square
Solve the equation 2x2 – 4x + 1 = 0 by completing
the square. Write the answer to 3 decimal places.
2x2 – 4x + 1 = 0
completing the square,
2(x – 1)2 – 1 = 0
2(x – 1)2 = 1
add 1 to both sides
(x – 1)2 =
divide both sides by 2,
x–1=±
square root both sides,
x=1+
x = 1.707
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1
2
1
2
or
1
2
x=1–
1
2
x = 0.293 (to 3 d. p)
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Contents
A6 Quadratic equations
A6.1 Solving quadratic equations by factorization
A6.2 Completing the square
A6.3 Using the quadratic formula
A6.4 Equations involving algebraic fractions
A6.5 Problems leading to quadratic equations
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Using the quadratic formula
Any quadratic equation of the form,
ax2 + bx + c = 0
can be solved by substituting the values of a, b and c into the
formula,
–b ± b2 – 4ac
x=
2a
This equation can be derived by completing the square on
the general form of the quadratic equation.
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Using the quadratic formula
Use the quadratic formula to solve x2 – 7x + 8 = 0.
1x2 – 7x + 8 = 0
–b ± b2 – 4ac
x=
2a
7 ± (–7)2 – (4 × 1 × 8)
x=
2×1
7 ± 49 – 32
x=
2
7 + 17
x=
2
x = 5.562
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or
7 – 17
x=
2
x = 1.438 (to 3 d.p.)
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Using the quadratic formula
Use the quadratic formula to solve 2x2 + 5x – 1 = 0.
2x2 + 5x – 1 = 0
–b ± b2 – 4ac
x=
2a
–5 ± 52 – (4 × 2 × –1)
x=
2×2
–5 ± 25 + 8
x=
4
–5 + 33
x=
4
x = 0.186
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or
–5 – 33
x=
4
x = –2.686 (to 3 d.p.)
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Using the quadratic formula
Use the quadratic formula to solve 9x2 – 12x + 4 = 0.
9x2 – 12x + 4 = 0
–b ± b2 – 4ac
x=
2a
12 ±  (–12)2 – (4 × 9 × 4)
x=
2×9
12 ± 144 – 144
x=
18
12 ± 0
x=
18
2
There is only one solution, x = 3
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Using the quadratic formula
Use the quadratic formula to solve x2 + x + 3 = 0.
1x2 + 1x + 3 = 0
–b ± b2 – 4ac
x=
2a
–1 ± 12 – (4 × 1 × 3)
x=
2×1
–1 ± 1 – 12
x=
2
–1 ± –11
x=
2
We cannot find –11 and so there are no solutions.
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Using b2 – 4ac
From using the quadratic formula,
–b ± b2 – 4ac
x=
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 positive, there are two solutions.
When b2 – 4ac is equal to zero, there is one solution.
When b2 – 4ac is negative, there are no solutions.
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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.
b2 – 4ac is positive
y
b2 – 4ac is zero
y
x
Two solutions
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b2 – 4ac is negative
y
x
One solution
x
No solutions
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Using b2 – 4ac
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Contents
A6 Quadratic equations
A6.1 Solving quadratic equations by factorization
A6.2 Completing the square
A6.3 Using the quadratic formula
A6.4 Equations involving algebraic fractions
A6.5 Problems leading to quadratic equations
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Equations involving algebraic fractions
Some equations involving algebraic fractions rearrange to
give quadratic equations. For example:
1
5
+
= 2
x
x+4
The first step when solving equations involving fractions is to
multiply through by the product of the denominators.
multiply by x(x + 4):
x + 4 + 5x = 2x(x + 4)
expand brackets and simplify: 6x + 4 = 2x2 + 8x
collect all terms on the r.h.s.:
0 = 2x2 + 2x – 4
divide by 2:
0 = x2 + x – 2
factorize:
0 = (x + 2)(x – 1)
x = –2
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or
x=1
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Equations involving algebraic fractions
Solve
4
3
–
= 1
x+2
x+8
Start by multiplying through by x + 2 and x + 8 the remove the
denominators,
4(x + 8) – 3(x + 2) = (x + 2)(x + 8)
expand the brackets: 4x + 32 – 3x – 6 = x2 + 10x + 16
x + 26 = x2 + 10x + 16
simplify:
collect all terms on the r.h.s.:
0 = x2 + 9x – 10
factorize:
0 = (x + 10)(x – 1)
x = –10
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or
x=1
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Equations involving algebraic fractions
Equations involving algebraic fractions may also lead to
quadratic equations that do not factorize. For example,
Solve,
multiply x(4 – x):
expand the brackets:
x
2
–
= 3
4–x
x
x2 + 2(4 – x) = 3x(4 – x)
x2 + 8 – 2x = 12x – 3x2
collect terms on the l.h.s.: 4x2 – 14x + 8 = 0
divide by 2:
2x2 – 7x + 4 = 0
This quadratic equation cannot be solved by factorization so
we have to solve it using the quadratic formula.
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Equations involving algebraic fractions
Using the quadratic formula to solve 2x2 – 7x + 4 = 0
–b ± b2 – 4ac
x=
2a
7 ± 72 – (4 × 2 × 4)
x=
2×2
7 ± 49 – 32
x=
4
7 + 17
x=
4
x = 2.781
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or
7 – 17
x=
4
x = 0.719 (to 3 d.p.)
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Contents
A6 Quadratic equations
A6.1 Solving quadratic equations by factorization
A6.2 Completing the square
A6.3 Using the quadratic formula
A6.4 Equations involving algebraic fractions
A6.5 Problems leading to quadratic equations
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Problems leading to quadratic equations
Some real-life problems can be solved using quadratic
equations. 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.
If her total journey time to work and back is 1 hour, what
was her average speed on the way to work?
distance
Remember, time taken =
average speed
Let Jenny’s average speed on the way to work be x.
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Problems leading to quadratic equations
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?
24
Jenny’s time taken to get to work =
x
24
Jenny’s time taken to get home from work =
x – 20
24
24
Total time there and back =
+
=1
x
x – 20
Solving this equation will give us the value of x, Jenny’s
average speed on the way to work.
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Problems leading to quadratic equations
24
24
+
=1
x
x – 20
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
48x – 480 = x2 – 20x
simplify:
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 solution is not possible in this situation?
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Problems leading to quadratic equations
The only solution that makes sense is x = 60 miles per hour.
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.
We can therefore ignore the second solution.
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.
This is usually because many physical quantities, such as
length, can only be positive.
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Problems leading to quadratic equations
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,
x+1
x–7
x
We can use Pythagoras’ Theorem to write an equation in
terms of x.
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Problems leading to quadratic equations
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:
collect on the l.h.s:,
factorize:
2x2 – 14x + 49 = x2 + 2x + 1
x2 – 16x + 48 = 0
(x – 4)(x – 12) = 0
x=4
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or
x = 12
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Problems leading to quadratic equations
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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