Solving Quadratic Equations (AQA GCSE Further Maths)
Revision Note
Written by: Mark Curtis
Reviewed by: Dan Finlay
Solving Quadratic Equations
How do I solve a quadratic equation using factorisation?
Factorise the quadratic and solve each bracket equal to zero
To solve
solve 2x – 3 = 0 to get x =
solve 3x + 5 = 0 to get x =
To solve don't forget to solve x = 0
The two solutions are x = 0 or x = 4
It is a common mistake to divide by x at the beginning (you will lose a solution)
How do I solve a quadratic equation by completing the square?
To solve x2 + bx + c = 0
replace the first two terms, x2 + bx, with (x + p)2 - p2 where p is half of b
this is called completing the square
x2 + bx + c = 0 becomes
rearrange this equation to make x the subject (using ±√)
For example, solve x2 + 10x + 9 = 0 by completing the square
x2 + 10x becomes (x + 5)2 - 52
so x2 + 10x + 9 = 0 becomes (x + 5)2 - 52 + 9 = 0
make x the subject (using ±√)
(x + 5)2 - 25 + 9 = 0
(x + 5)2 = 16
x + 5 = ±√16
x = ±4 - 5
x = -1 or x = -9
If the equation is ax2 + bx + c = 0 with a number in front of x2, then divide both sides by a first, before completing the square
How do I use the quadratic formula to solve a quadratic equation?
The quadratic formula is
Read off the values of a, b and c from the equation
Substitute these into the formula
Write this line of working in the exam
Put brackets around any negative numbers being substituted in
To solve 2x2 - 7x - 3 = 0 using the quadratic formula:
a = 2, b = -7 and c = -3
Type this into a calculator
once with + for ± and once with - for ±
The solutions are x = 3.886 and x = -0.386 (to 3 dp)
Rounding is often asked for in the question
The calculator also gives these solutions in exact form (surd form)
x = and x =
You need to be able to find solutions in exact / surd form without a calculator
this means working out (-7)2 - 4 × 2 × (-3)
What is the discriminant?
The part of the formula under the square root (b2 – 4ac) is called the discriminant
The sign of this value tells you if there are 0, 1 or 2 solutions
If b2 – 4ac > 0 (positive)
then there are 2 different solutions
If b2 – 4ac = 0 (zero)
then there is only 1 solution
sometimes called "repeated solutions"
If b2 – 4ac < 0 (negative)
then there are no solutions
If your calculator gives you solutions with i terms in, these are "complex" and not what we are looking for
Interestingly, if b2 – 4ac is a perfect square number ( 1, 4, 9, 16, …) then the quadratic expression could have been factorised!
How does completing the square link to the quadratic formula?
The quadratic formula actually comes from completing the square of ax2 + bx + c = 0
You can see hints of this when you solve quadratics
For example, solving x2 + 10x + 9 = 0
by completing the square, (x + 5)2 = 16 so x = -5 ± 4
by the quadratic formula, = -5 ± 4
Can I use my calculator to solve quadratic equations?
Yes, in the calculator paper, use a calculator to check your final solutions!
Calculators also help you to factorise (if you're struggling with that step)
A calculator gives solutions to as x = and x =
"Reverse" the method above to factorise
Warning: a calculator gives solutions to 12x2 + 2x – 4 = 0 as x = and x =
But 12x2 + 2x – 4
the right-hand side expands to 6x2 + ... ,not 12x2 + ...
Correct this by multiplying the right by 2
12x2 + 2x – 4
Examiner Tips and Tricks
Make sure the quadratic equation has "= 0" on the right-hand side, otherwise it needs rearranging first
rearrange to have ax2 on its positive side (a>0)
Always look for how the question wants you to leave your final answers
2 decimal places, 3 significant figures, in exact form, etc
Worked Example
(a) Solve , giving your answers in exact form.
“exact form” suggests using the quadratic formula (surds will be in the answer)
Substitute a = 1, b = -7 and c = 2 into the formula, putting brackets around any negative numbers
Work out (-7)2 - 4 × 1 × 2 and simplify
This is as simplified as possible
(b) Solve
Method 1
If you cannot spot the factorisation and this is in the calculator paper, use the quadratic formula
Substitute a = 16, b = -82 and c = 45 into the formula, putting brackets around any negative numbers
Use a calculator to find each solution
or
Method 2
If you do spot the factorisation, (2x – 9)(8x – 5), then use that method instead
Set the first bracket equal to zero
Add 9 to both sides then divide by 2
Set the second bracket equal to zero
Add 5 to both sides then divide by 8
or
(c) By writing in the form , solve
This question wants you to complete the square first
Find p (by halving the middle number)
Write x2 + 6x as (x + p)2 - p2
Replace x2 + 6x with (x + 3)2 – 9 in the equation
Make x the subject of the equation (start by adding 4 to both sides)
Take square roots of both sides (include a ± sign to get both solutions)
Subtract 3 from both sides
Find each solution separately using + first, then - second
Even though the quadratic factorises to (x + 5)(x + 1), this is not the method asked for in the question
Hidden Quadratic Equations
How do I spot a hidden quadratic equation?
Hidden quadratics have the same structure as quadratic equations
a(something)2 + b(something) + c = 0
Here are some hidden quadratics based on x2 - 3x - 4 = 0:
(a quadratic in x2)
(a quadratic in x8)
(a quadratic in because is )
(a quadratic in because )
Sometimes, a change of base helps to spot a hidden quadratic
e.g. the first term in can be written
is a quadratic in
How do I solve a hidden quadratic equation?
You can solve a(...)2 + b(...) + c = 0 with a substitution
Substitute "u = ..." and rewrite the equation in terms of u only
au2 + bu + c = 0
Solve this easier quadratic equation in u to get u = p and u = q
Replace the u's with their substitution to get two equations
"... = p" and "... = q"
Solve these two separate equations to find all the solutions
These equations might have multiple solutions or none at all!
e.g. to solve x4 - 3x2 - 4 = 0
Substitute u = x2 to get u2 - 3u - 4 = 0
The solutions are u = 4 or u = -1,
Rewrite in terms of x:
x2 = 4 or x2 = -1,
Solve to give x = -2 or x = 2 (no solutions from x2 = -1 as you can't square-root a negative)
Examiner Tips and Tricks
While the substitution method is not compulsory, beware of skipping steps
e.g. it is incorrect to "jump" from the solutions of x2 -3x - 4 = 0 to the solutions of (x + 5)2 - 3(x + 5) - 4 = 0 by "adding 5 to them"
the substitution method shows you end up subtracting 5
Worked Example
(a) Solve
This is a quadratic in x4 so let u = x4
Solve this simpler quadratic equation, for example by factorisation
Write out the u solutions
Replace u with x4
Solve these separate equations (remember an even power gives two solutions)
Write your solutions out (it's good practice to write them in numerical order)
(b) Solve
This is a quadratic in √x so let u = √x
Solve this simpler quadratic equation, for example by factorisation
Replace u with √x and solve
You can check your solutions by substituting them back into the equation
If you put x = 4 as a solution by mistake then substituting will spot this error
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