Quadratic Equation Methods (CIE IGCSE Additional Maths)

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

Solve

x^{2} - 7 x + 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

16 x^{2} - 82 x + 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

$2 x - 9 = 0$

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

$8 x - 5 = 0$

Add 5 to both sides then divide by 8

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

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

(c)

By writing

x^{2} + 6 x + 5

in the form

{(x + p)}^{2} + q

, solve

x^{2} + 6 x + 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 + 6x as (x + p)² - p²

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

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

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

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)

$x + 3 = \pm \sqrt{4} = \pm 2$

Subtract 3 from both sides

$x = \pm 2 - 3$

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:
- $x^{4} - 3 x^{2} - 4 = 0$ (a quadratic in x²)
- $x^{16} - 3 x^{8} - 4 = 0$ (a quadratic in x⁸)
- $x - 3 \sqrt{x} - 4 = 0$ (a quadratic in $\sqrt{x}$ because ${(\sqrt{x})}^{2}$ is $x$ )
- $x^{\frac{2}{3}} - 3 x^{\frac{1}{3}} - 4 = 0$ (a quadratic in $x^{\frac{1}{3}}$ because ${(x^{\frac{1}{3}})}^{2} = x^{\frac{2}{3}}$ )
Sometimes, a change of base helps to spot a hidden quadratic
- e.g. the first term in $4^{x} - 3 \times 2^{x} - 4 = 0$ 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. $3 \tan^{2} 3 x + 4 \tan 3 x - 6 = 0$ is a quadratic in $\tan 3 x$

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)

Exam 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 $x^{8} - 17 x^{4} + 16 = 0$

This is a quadratic in x⁴ so let u = x⁴

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

Solve this simpler quadratic equation, for example by factorisation

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

Write out the u solutions

$u = 16 or u = 1$

Replace u with x⁴

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

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

$x = \pm 2 or x = \pm 1$

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

$x = - 2, - 1, 1 or 2$

(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

$(u - 3) (u + 2) = 0 u = 3 or u = - 2$

Replace u with √x and solve

$\sqrt{x} = 3 \Rightarrow x = 9 \sqrt{x} = - 2 has no solutions as \sqrt{x} \geq 0$

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

$9 - \sqrt{9} - 6 = 0 4 - \sqrt{4} - 6 = - 4 \neq 0$

$x = 9$

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