Download GCSE Session 13 – The Quadratic Formula

Session 13 - The Quadratic Formula
Quadratics ax2 + bx + c

Factorise using 2 sets of brackets

(ax + p) (x + q)

The numbers p and q must still multiply
together to give c, only this time, one of
them must be multiplied by a then added
to the other to give b.

Example – factorise 2x2 + 11x + 15

(2x + p) (x + q)

The factors of 15 are
15 and 1
3 and 5
Now we are looking for the pair of factors that, after one of
them is multiples by 2, they add together to give 11





15 + (2 x 1) =
1 + (2 x 15) =
3 + (2 x 5) =
5 + (2 x 3) =
17
31
12
11

Here the 3 must be multiplied by the 2, so they must be in
different brackets. (2x + 5) (x + 3)

Ex19.3 Q2 (discuss negatives eg

Example – factorise 4x2 + 4x + 1

(4x + p) (x + q)

The factors of 1 are
1 and 1 0r -1 and -1

Now we are looking for the pair of factors that, after
one of them is multiples by 4, they add together to
give 4

1 + (4 x 1) = 5
-1 + (4 x -1) = -5


is this the only option?
Neither of these work, so is there another factor we
can take out? A different way of setting out the
brackets?
Examples with negatives

Example – factorise 2y2 - 9y + 7

(2y + p) (x + q)

The factors of 7 are
1 and 7

Now we are looking for the pair of factors that,
after one of them is multiplied by 2, they add
together to give 9



1 + (2 x 7) = 15
7 + (2 x 1) = 9
(2y 7) (y 1) what signs are needed?
Solving Quadratics
Once we have factorised, we need to find the values for
x which will bring the bracket to 0. this will give us
results for x and solve the equation.
Eg
(x - 3)(x + 2) = 0
In order to make the whole expression come to zero, we
need to make each bracket (in turn) come to zero.
For the first bracket to be zero. X = 3
For the second bracket to be zero. X = -2
So the solutions for this equation are 3 and -2 (this is
where the curve would cross the x axis.)
 The coordinates of the x intersect are (-2,0) and (3,0)

Ex 19.5 Q1
Extension some of Q2
Solving Quadratics

Some quadratics expressions will not
factorise, these can still be solved.

We can use a graph as we have already
seen, but these are unreliable when we
have to sketch in the curve.

There is a formula that will always work.
As long as we can arrange the quadratic
into the form:
ax2 + bx + c = 0
The quadratic formula is given to you in the
exam paper (higher only)
Using the quadratic formula
Writ the formula out
 List a b and c
x2 - 5x + 2 = 0
a=1
b = -5
c=2
Now substitute these values into the
equation. Once you’ve simplified it as far
as possible, you must separate the plus or
minus to give you 2 different equations.

This will generate 2 different answers for x
X = 0.43844… and X = 4.5615….
Exercise 19.6 Q1 and a few of Q2
 Extension Q5

Completing the square

A perfect square would be when a
factorised quadratic has 2 identical
brackets
(x+1)(x+1) = (x+1) 2

Other expressions that are not prefect
square can be written in this form, but need
something adding on after the brackets are
squared

E.g.
(x+1) 2 + 3

To write ax2 + bx + c in completed square
form. Square term is
(x + ½ coefficient of x) 2
 You then multiply this out, and adjust for the
constant value

Your answer needs to be in the form:
 (x + a) 2 + b


Where a is half the coefficient of x from the
original equation, and b is the adjustment.
Write : x 2 + 6x + 10 in the form (x + a) 2 + b
i.e. completed square form:
We have a squared bracket that contains x, plus half
the coefficient of x. The coefficient of x in this
expression is 6, so half that is 3

(x + 3) 2 if we expand the bracket we get:
x2 + 3x + 3x + 9
=
x 2 + 6x + 9
We have +9 on the end, we need +10, so the
adjustment is to add 1
(x + 3) 2 + 1
Ex 19.7 Q3
Using completing the square to
solve quadratics
x 2 + 2x – 8 = 0 can be re written as:
x 2 + 2x = 8
Now we can rewrite the LHS if the equation in
completed square form
(x + a) 2 + b
(x + 1) 2 – 1 = 8
(x + 1) 2 = 9
Now take the square root of both sides
x + 1 = ± √9
So
So
x+1=3
x=2
and
and
x + 1 = -3
x = -4
Ex19.8 Q1
Problem solving

You need to create quadratic equations
with only one variable (usually x)
Ex 19.9 Q 6
 Extension Q4 and 9


The context of the question may affect
whether you need 2 answers or only
positive answers.
Class Discussion p158

Graphs of reciprocals, exponential and
circles.

You need to be able to recognise and
sketch these basic shapes of graphs,
and link them to the appropriate formula

We will revisit this topic during revision

Trial and improvement – higher learners
only

Sometimes there’s no way (at GCSE level)
to find the solution to equations, so
systematic trial and improvement can be
used (not for quadratics) Pg 207

For finding the solutions to cubics, you can
substitute in a guess, the systematically
adjust your guess until you have reached
then required degree of accuracy. Ex21.6
Q1