Solving Quadratic Equations (AQA GCSE Further Maths)

Revision Note

Mark Curtis

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 open parentheses 2 x minus 3 close parentheses open parentheses 3 x plus 5 close parentheses equals 0

    • solve 2x – 3 = 0 to get x3 over 2

    • solve 3x + 5 = 0 to get xnegative 5 over 3

  • To solve x open parentheses x minus 4 close parentheses equals 0 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 

    • x equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction

  • 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

    • x equals fraction numerator negative open parentheses negative 7 close parentheses plus-or-minus square root of open parentheses negative 7 close parentheses squared minus 4 cross times 2 cross times open parentheses negative 3 close parentheses end root over denominator 2 cross times 2 end fraction

    • 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)

    • xfraction numerator 7 plus square root of 73 over denominator 4 end fraction and xfraction numerator 7 minus square root of 73 over denominator 4 end fraction

    • 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!

  • 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,  x equals fraction numerator negative 10 plus-or-minus square root of 64 over denominator 2 end fraction equals negative 5 plus-or-minus 8 over 2 = -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 6 x squared plus x minus 2 equals 0 as xnegative 2 over 3  and x1 half

    • "Reverse" the method above to factorise

      • 6 x squared plus x minus 2 identical to open parentheses 3 x space plus space 2 close parentheses open parentheses 2 x space minus space 1 close parentheses

    • Warning: a calculator gives solutions to 12x2 + 2x – 4 = 0 as xnegative 2 over 3 and x1 half 

      • But 12x2 + 2x – 4 not identical to open parentheses 3 x plus 2 close parentheses open parentheses 2 x minus 1 close parentheses

      • the right-hand side expands to 6x2 + ... ,not 12x2 + ...

      • Correct this by multiplying the right by 2

      • 12x2 + 2x – 4 identical to 2 open parentheses 3 x plus 2 close parentheses open parentheses 2 x minus 1 close parentheses

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 x squared minus 7 x plus 2 equals 0, 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

  x equals fraction numerator negative open parentheses negative 7 close parentheses plus-or-minus square root of open parentheses negative 7 close parentheses squared minus 4 cross times 1 cross times 2 end root over denominator 2 cross times 1 end fraction

Work out (-7)2 - 4 × 1 × 2 and simplify

x equals fraction numerator 7 plus-or-minus square root of 41 over denominator 2 end fraction  

This is as simplified as possible

bold italic x bold equals fraction numerator bold 7 bold plus-or-minus square root of bold 41 over denominator bold 2 end fraction

(b) Solve 16 x squared minus 82 x plus 45 equals 0
 

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

x equals fraction numerator negative open parentheses negative 82 close parentheses plus-or-minus square root of open parentheses negative 82 close parentheses squared minus 4 cross times 16 cross times 45 end root over denominator 2 cross times 16 end fraction

Use a calculator to find each solution

 x equals 9 over 2or x equals 5 over 8

Method 2
If you do spot the factorisation, (2x – 9)(8x – 5), then use that method instead

open parentheses 2 x minus 9 close parentheses open parentheses 8 x minus 5 close parentheses equals 0 

Set the first bracket equal to zero

2 x minus 9 equals 0 

Add 9 to both sides then divide by 2

table row cell 2 x end cell equals 9 row x equals cell 9 over 2 end cell end table

Set the second bracket equal to zero

8 x minus 5 equals 0 

Add 5 to both sides then divide by 8

table row cell 8 x end cell equals 5 row x equals cell 5 over 8 end cell end table

x equals 9 over 2or x equals 5 over 8

 

(c) By writing x squared plus 6 x plus 5 in the form open parentheses x plus p close parentheses squared plus q, solve x squared plus 6 x plus 5 equals 0
 

This question wants you to complete the square first
Find p (by halving the middle number)

p equals 6 over 2 equals 3 

Write x2 + 6x as (x + p)2 - p2

table row cell x squared plus 6 x end cell equals cell open parentheses x plus 3 close parentheses squared minus 3 squared end cell row blank equals cell open parentheses x plus 3 close parentheses squared minus 9 end cell end table 

Replace x2 + 6x with (x + 3)2 – 9 in the equation

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x plus 3 close parentheses squared minus 9 plus 5 end cell equals 0 row cell open parentheses x plus 3 close parentheses squared minus 4 end cell equals 0 end table

Make x the subject of the equation (start by adding 4 to both sides)

open parentheses x plus 3 close parentheses squared equals 4 

Take square roots of both sides (include a ± sign to get both solutions)

x plus 3 equals plus-or-minus square root of 4 equals plus-or-minus 2 

Subtract 3 from both sides

x equals plus-or-minus 2 minus 3 

Find each solution separately using + first, then - second

x equals negative 5 comma space x equals negative 1

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:

    • x to the power of 4 minus 3 x squared minus 4 equals 0 (a quadratic in x2)

    • x to the power of 16 minus 3 x to the power of 8 minus 4 equals 0 (a quadratic in x8)

    • x minus 3 square root of x minus 4 equals 0 (a quadratic in square root of x because open parentheses square root of x close parentheses squared is x)

    • x to the power of 2 over 3 end exponent minus 3 x to the power of 1 third end exponent minus 4 equals 0 (a quadratic in x to the power of 1 third end exponentbecause open parentheses x to the power of 1 third end exponent close parentheses squared equals x to the power of 2 over 3 end exponent)

  • Sometimes, a change of base helps to spot a hidden quadratic

    • e.g. the first term in 4 to the power of x minus 3 cross times 2 to the power of x minus 4 equals 0 can be written 4 to the power of x equals open parentheses 2 squared close parentheses to the power of x equals 2 to the power of 2 x end exponent equals open parentheses 2 to the power of x close parentheses squared

      • open parentheses 2 to the power of x close parentheses squared minus 3 cross times 2 to the power of x minus 4 equals 0 is a quadratic in 2 to the power of x

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  x to the power of 8 minus 17 x to the power of 4 plus 16 equals 0

This is a quadratic in x4 so let u = x4

u squared minus 17 u plus 16 equals 0

Solve this simpler quadratic equation, for example by factorisation

open parentheses u minus 16 close parentheses open parentheses u minus 1 close parentheses equals 0 

Write out the u solutions

u equals 16 space space space or space space space u space equals space 1

Replace u with x4

x to the power of 4 equals 16 space space space or space space space x to the power of 4 equals 1

Solve these separate equations (remember an even power gives two solutions)

x equals plus-or-minus 2 space space or space space x equals plus-or-minus 1

Write your solutions out (it's good practice to write them in numerical order) 

bold italic x bold equals bold minus bold 2 bold comma bold space bold minus bold 1 bold comma bold space bold 1 bold space bold or bold space bold 2

(b) Solve  x minus square root of x minus 6 equals 0

This is a quadratic in √x so let u = √x

u squared minus u minus 6 equals 0

Solve this simpler quadratic equation, for example by factorisation

open parentheses u minus 3 close parentheses open parentheses u plus 2 close parentheses equals 0
u equals 3 space or space u equals negative 2

Replace u with √x and solve

square root of x equals 3 rightwards double arrow x equals 9
square root of x equals negative 2 space has space no space solutions space as space square root of x greater or equal than 0

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

9 minus square root of 9 minus 6 equals 0
4 minus square root of 4 minus 6 equals negative 4 not equal to 0

bold italic x bold equals bold 9

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Mark Curtis

Author: Mark Curtis

Expertise: Maths

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.