Quadratic Equation Methods (Cambridge International IGCSE Further Maths)

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. $x = \frac{- 4 \pm \sqrt{4^{2} - 4 \times 2 \times ( - 2 )} }{2 \times 2} = \frac{- 4 \pm \sqrt{32} }{4} = \frac{- 4 \pm 4 \sqrt{2} }{4} = - 1 \pm \sqrt{2}$

• 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

Exam Hint

• 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 $x^{2}-7x+2=0$, 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

$x = \frac{- ( - 7 ) \pm \sqrt{( - 7 )^{2} - 4 \times 1 \times 2} }{2 \times 1}$

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 $16x^{2}-82x+45=0$

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

$x = \frac{- ( - 82 ) \pm \sqrt{( - 82 )^{2} - 4 \times 16 \times 45} }{2 \times 16}$

Use a calculator to find each solution

x = $\frac{9}{2}$or x =$\frac{5}{8}$

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

$16 x^{2} - 82 x + 45 = ( 2 x - 9 ) ( 8 x - 5 ) = 0$

Set the first bracket equal to zero

$2x-9=0$

Add 9 to both sides then divide by 2

$\begin{array}{rcl} 2 x & = 9 \\ x = \frac{9}{2} \end{array}$

Set the second bracket equal to zero

$8x-5=0$

Add 5 to both sides then divide by 8

$\begin{array}{rcl} 8 x & = 5 \\ x = \frac{5}{8} \end{array}$

Final answer: x = $\frac{9}{2}$or x =$\frac{5}{8}$

(c) By writing $x^{2}+6x+5$ in the form $( x + p )^{2} + q$, solve $x^{2}+6x+5=0$

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)

$p=\frac{6}{2}=3$

Write x² + 6**x **as (x + p)²* -** p*^{2 * *}

$\begin{gathered} x^{2} + 6 x & = ( x + 3 )^{2} - 3^{2} \\ = ( x + 3 )^{2} - 9 \end{gathered}$

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

$\begin{gathered} ( x + 3 )^{2} - 9 + 5 & = 0 \\ ( x + 3 )^{2} - 4 & = 0 \end{gathered}$

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

$( x + 3 )^{2} = 4$

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

$x+3=\pm\sqrt{4}=\pm2$

Subtract 3 from both sides

$x=\pm2-3$

Find each solution separately using + first, then - second

Final answer: 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:

  • $x^{4}-3x^{2}-4=0$ (a quadratic in x²)
  • $x^{16}-3x^{8}-4=0$ (a quadratic in x⁸)
  • $x-3\sqrt{x}-4=0$ (a quadratic in $\sqrt{x}$ because $( \sqrt{x} )^{2}$ is $x$)
  • $x^{\frac{2}{3}}-3x^{\frac{1}{3}}-4=0$ (a quadratic in $x^{\frac{1}{3}}$because $( x^{1 \frac{1}{3}} )^{2} = x^{ \frac{2}{3} }$)

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

  • e.g. the first term in $4^{x}-3\times2^{x}-4=0$ can be written $4^{x = ( 2^{2} )^{x} = 2^{2} x} = ( 2^{x} )^{2}$

    • $( 2^{x} )^{2} - 3 \times 2^{x} - 4 = 0$ is a quadratic in $2^{x}$

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

  • e.g. $3\mathrm{tan}^{2}3x+4\mathrm{tan}3x-6=0$ is a quadratic in $\mathrm{tan}3x$

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)

Exam Hint

• 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 $x^{8}-17x^{4}+16=0$

This is a quadratic in x⁴ so let u =* x*⁴

$u^{2}-17u+16=0$

Solve this simpler quadratic equation, for example by factorisation

$( u - 16 ) ( u - 1 ) = 0$

Write out the u solutions

$u=16\mathrm{or}u=1$

Replace u with x⁴

$x^{4}=16\mathrm{or}x^{4}=1$

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

$x=\pm2\mathrm{or}x=\pm1$

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

$x=-2,-1,1\mathrm{or}2$

(b) Solve $x-\sqrt{x}-6=0$

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

$u^{2}-u-6=0$

Solve this simpler quadratic equation, for example by factorisation

$( u - 3 ) ( u + 2 ) = 0 u = 3 \text{or} u = - 2$

Replace u with √x and solve

$\sqrt{x}=3\Rightarrow x=9\,\sqrt{x}=-2\mathrm{has} \mathrm{no} \mathrm{solutions} \mathrm{as}\sqrt{x}\geq0$

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

$9-\sqrt{9}-6=0\,4-\sqrt{4}-6=-4\neq0$

$x=9$