Mr Barton’s Maths Notes
Algebra
7. Solving Quadratic
Equations
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7. Solving Quadratic Equations
The three ways to solve quadratic equations…
Again, it will depend on your age and maths set as to how many of these you need to know,
but here are the three ways which we can solve quadratic equations:
1. Factorising
2. Using the Quadratic Formula
3. Completing the Square
Which ever way you choose (or are told to do!), you must remember the Golden Rule:
The Golden Rule for Solving Quadratic Equations: You should always get TWO answers…
Note: in actual fact one (or even both) answers may not exist, but you don’t need to worry
about that until A Level!
Why on earth do I get two answers?…
This is all to do with the fact that quadratics contain squares, and what happens when we
square negative numbers…
Imagine you were trying to think solve this equation:
x2  25
Well, x = 5 is definitely a solution that works, but there is another…erm… erm…
What about x = -5!... Because when you square a negative number, you get a positive answer!
And that’s why we get two solutions when quadratics are involved!
1. Solving by Factorising
This is by far the easiest and quickest way to solve a quadratic equation, and if you are not
told otherwise, then always spend a minute or so seeing if the equation will factorise.
Note: For the rest of this section, I am going to assume you are comfortable with what was
covered in Algebra 6. More Factorising. Please go back and have a quick read if not.
Method
1. Re-arrange the equation to make it equal to zero
2. Factorise the quadratic equation
3. Think what value of the unknown letter would make each of your brackets equal to zero
4. These two numbers are your answers!
Why on earth does that work?
Imagine, after following steps 1. and 2., you find yourself looking at this…
( x  4) ( x  3)  0
Think about what we have got here… we have two things (x – 4) and (x + 3) that when
multiplied together (disguised multiplication sign between the brackets) equal zero
Well… if two things multiplied together equal zero, then at least one of them must be zero!
So… you ask yourself: “what value of x makes the first bracket equal to zero?”... 4!
And… “what value of x makes the second bracket equal to zero?”... -3!
So we have our answers:
x  4
or
x  3
Example 1
Example 2
x2  3x  28  0
Okay, let’s go through each stage of the method:
1. The equation is already equal to zero, so that is
a bonus!
2. Let’s factorise the left hand side, like the good
old days…
x  3x  28  ( x  7) ( x  4)
2
And so, in terms of our equation, we have:
( x  7) ( x  4)  0
3. Right, we need to pick some values of x to make
each of the brackets equal to zero:
( x  7) ( x  4)  0
x  7
x  4
4. So, we have our answers…
x  7
or
2 x2  5x  3
1. Problem: the equation is NOT equal to zero… but
if we subtract 3 from both sides, we’re good to go!
2 x2  5x  3  0
2. This is one of the tricky factorisations…
2x2  5x  3  (2x  1) ( x  3)
And so, in terms of our equation, we have:
(2 x  1) ( x  3)  0
3. Right, we need to pick some values of x to make
each of the brackets equal to zero:
(2 x  1) ( x  3)  0
x  1
2
x  3
4. So, we have our answers…
x  4
x  1
2
or
x  3
2. Solving by using the Quadratic Formula
The good news is that the quadratic formula can solve every single quadratic equations
The bad news is that it looks complicated and it’s fiddly to use!
+ or – this is where
the 2 answers
come from
The Quadratic Formula:
x 
b 
b2  4ac
2a
Note: To be able to use this
formula, you must be very good at
using your calculator. Practice to
make sure you can get the
answers I get below, and if not
then ask you teacher!
What do the letters stand for?...
The letters are just the coefficients (the numbers in front of) the unknowns in your equation:
Remember: as always, you must include the signs of the numbers as well!
ax + bx + c  0
2
Example
5x2  8x + 12  0
a  5
b  8
c  12
Never Ever Forget: Before you start sticking numbers into the formula, you must make sure
that you rearrange your equation to make it equal to zero!
Example 1
Pressing the buttons on the calculator
You’ll be amazed how many people throw away
easy marks because they can’t use their
calculator properly!
x  4x  2  0
2
What a nice looking equation. I bet it
factorises… erm… erm… no it doesn’t!
Here is one order of buttons you could press to
get you the correct answer!
So we’ll have to use the formula.
It’s already equal to zero, so we just need to
figure out what our a, b and c are:
ax + bx + c  0
2
x2  4 x  2  0
a  1
b  4
Top Tip: always put any
negative numbers in
brackets or calculators
tend to do daft things!
c  2
Note: a = 1, and not 0! Remember, the 1 is hidden!
Stick the numbers in our formula…
x 
4 
6.828427125
This gives you the value of the top of the fraction
42  4 1 2
2 1
And if you are careful with your calculator, you
should get…
x  3.41 or x  0.59
Change this to
minus to work out
the 2nd answer!
(2dp)
3.414213562
Changing the sign to minus gives the other
answer of…
0.585786437
Example 2
Pressing the buttons on the calculator
5x2  10  3x
It’s not going to factorise, and it’s not equal to
zero!
Okay, this time we’ll work out the answer that
uses the minus on top of the fraction instead of
the plus…
So before we use the formula we must… add 3x
and subtract 10 from both sides to give us:
5x2  3x  10  0
Top Tip: always put any
negative numbers in
brackets or calculators
tend to do daft things!
ax + bx + c  0
2
5x2  3x  10  0
a  5
b  3
c   10
-17.45683229
Stick the numbers in our formula…
x 
3 
3  4  5  10
25
2
And if you are careful with your calculator, you
should get…
x  1.15 or
Change this to plus
to work out the 2nd
answer!
x   1.75
(2dp)
This gives you the value of the top of the fraction
-1.745683229
Changing the sign to plus gives the other
answer of…
1.145683229
3. Solving by Completing the Square
How would you factorise this?...
x 2  10 x
It doesn’t look like it can be done, but what about if I write it like this…
( x  5)2
Now that is certainly factorised as it is in brackets, but is it the correct answer?…
Let’s expand the brackets using FOIL to find out…
( x  5)2  ( x  5) ( x  5)  x2  5x  5x  25  x2  10x  25
It’s close! In fact, our factorised version is just 25 too big! So, we can say…
x2  10x  ( x  5)2  25
And that is completing the square… the square is the (x + 5)2, and the - 25 completes it!
Method for Completing the Square
1. If the number in front of the x2 is NOT 1, then take out a factor to make it so
2. Complete the Square using this fancy looking formula:
x  bx  ( x  b )  ( b )
2
2
2
2
2
Note: b, is just the number (with
sign!) in front of the x, like 10 in the
example above!
3. If you need to solve the equation, use SURDS … Crucial: When you square root, you must
take both the positive and the negative to make sure you get TWO answers!
Example 1 Complete the square and solve:
x  4 x  21
Example 2 Complete the square and solve:
4 x2  8x  21
2
1. The number in front of
x2
is 1, so we’re fine!
1. The number in front of x2 is 4, so we must take
out a factor of 4 to sort things out!
2. Let’s use the formula on the left hand side:
4 ( x2  2 x)  21
x 2  bx  ( x  b ) 2  ( b ) 2
2
2
2. Use the formula on the terms in the brackets:
x 2  4 x  ( x  4 )2  (4 )2
2
2
x 2  2 x  ( x  2 )2  (2 )2
2
2
( x  2)  4
( x  1)2  1
2
So now we have:
( x  2)  4  21
2
3. Well, so long as you are good at solving
equations, you’ll be fine from here…
+4
√
+2
So…
( x  2)2  25
So now we have:
Expanding:
4 ( x  1)2  4  21
3. Time to solve…
+4
x  2   25   5
÷4
x  5  2
√
Or
x  5  2  7
x  5  2  3
4 ( x  1) 2  1  21
+1
4( x  1)2  25
( x  1)2  6.25
x  1   6.25   2.5
x   2.5  1
x  2.5  1  3.5
Or
x   2.5  1   1.5
Good luck with
your revision!