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Quadratic Equation Methods (CIE IGCSE Additional Maths)

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Mark

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Mark

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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 Tip

  • 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 Tip

  • 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

    Author: Mark

    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.