Revision NotesExam QuestionsPast Papers
O LevelFurther MathsCambridge (CIE)Additional MathsRevision NotesAlgebra & FunctionsQuadratic FunctionsQuadratic Equation Methods

Quadratic Equation Methods (Cambridge (CIE) O Level Additional Maths): Revision Note

Last updated

Quadratic Equation Methods

If you have to solve a quadratic equation but are not told which method to use, here is a guide as to what to do

When should I solve by factorisation?

  • When the question asks to solve by factorisation

    • For example, part (a) Factorise 6x2 + 7x – 3, part (b) Solve  6x2 + 7x – 3 = 0

  • When solving two-term quadratic equations

    • For example, solve x2 – 4x = 0

      • …by taking out a common factor of x to get x(x – 4) = 0

      • ...giving x = 0 and x = 4

    • For example, solve x2 – 9 = 0

      • …using the difference of two squares to factorise it as (x + 3)(x – 3) = 0

      • ...giving x = -3 and x = 3

      • (Or by rearranging to x2 = 9 and using ±√ to get x =  = ±3)

  • When possible, factorising is usually the easiest way to solve a quadratic equation

    • Even on the calculator paper, if you can spot a factorisation quickly, use this approach

When should I use the quadratic formula?

  • If the coefficients (a, b and c) are large, factorising and completing the square can be difficult or slow

    • The quadratic formula lends itself to using a calculator

    • Some modern calculators will solve quadratic equations directly, with no need to use the formula

  • Typically the quadratic formula would be used when rounding is involved

    • For example, if a question says to leave solutions correct to 2 decimal places or 3 significant figures

  • However, the quadratic formula is also useful when answers need to be exact

    • The formula lends itself to surd form after simplifying some of the values within it

    • e.g.  x equals fraction numerator negative 4 plus-or-minus square root of 4 squared minus 4 cross times 2 cross times open parentheses negative 2 close parentheses end root over denominator 2 cross times 2 end fraction equals fraction numerator negative 4 plus-or-minus square root of 32 over denominator 4 end fraction equals fraction numerator negative 4 plus-or-minus 4 square root of 2 over denominator 4 end fraction equals negative 1 plus-or-minus square root of 2

  • If in doubt, use the quadratic formula - it always works

When should I solve by completing the square?

  • A question may direct you to solve by completing the square

    • e.g.  Part (a) says to complete the square and part (b) says 'hence' or 'use part (a)' to solve ...

  • Completing the square may have already happened for other reasons

    • e.g.  Completing the square allows the coordinates of the turning point on a quadratic graph to be found easily

    • If this has been done in an earlier part of a question, use it to solve the quadratic equation

Examiner Tips and Tricks

  • Calculators can solve quadratic equations

    • Double check you've entered the equation correctly, in the correct format

    • Use this feature to check your answers where possible

    • If the solutions on your calculator are whole numbers or fractions (with no square roots), this means the quadratic equation does factorise

Worked Example

a) Solve x squared minus 7 x plus 2 equals 0, giving your answers correct to 2 decimal places
 

“Correct to 2 decimal places” suggests using the quadratic formula and/or a calculator
For accuracy, it is a good idea to use both - use the formula and calculator as normal first
Then use the quadratic solver feature to check your solutions

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

Use a calculator to find each solution
 

x = 6.70156… or 0.2984...
 

Round your final answers to 2 decimal places

x = 6.70 or x = 0.30

If your calculator has a quadratic equation solver, use it to check your answers

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

Method 1
The coefficients are large and so the factorisation, even if possible, is hard to spot
Therefore, one method to use is the quadratic formula - it always works!
The solution below is the manual way to use a calculator, but as above, if your calculator has a quadratic solver feature, you may use that
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 = bold 9 over bold 2  or x = bold 5 over bold 8

Method 2
If you do persevere with the factorisation then use that method instead
 

16 x squared minus 82 x plus 45 equals 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 attributes columnalign right center left columnspacing 0px end attributes 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 attributes columnalign right center left columnspacing 0px end attributes row cell 8 x end cell equals 5 row x equals cell 5 over 8 end cell end table

x = bold 9 over bold 2  or x = bold 5 over bold 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
 

Notice this question does not use the phrase 'completing the square' but shows the form of it instead
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 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 = - 5, x = - 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

  • Trigonometric equations can also be in the form of a quadratic

    • e.g. 3 tan squared space 3 x plus 4 tan space 3 x minus 6 equals 0 is a quadratic in tan space 3 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

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

    Test yourself

    Did this page help you?

    Previous:Completing the SquareNext:Discriminants