Solving Quadratic Equations (AQA GCSE Further Maths)

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Solving Quadratic Equations

How do I solve a quadratic equation using factorisation?

Factorise the quadratic and solve each bracket equal to zero
To solve $(2 x - 3) (3 x + 5) = 0$
- solve 2x – 3 = 0 to get x = $\frac{3}{2}$
- solve 3x + 5 = 0 to get x = $- \frac{5}{3}$
To solve $x (x - 4) = 0$ 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
    - (x + p)² - p² + c = 0 where p is half of b
- 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
- $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

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
- $x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 \times 2 \times (- 3)}}{2 \times 2}$
- 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)
- - x = $\frac{7 + \sqrt{73}}{4}$ and x = $\frac{7 - \sqrt{73}}{4}$
- 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!

How does completing the square link to the quadratic formula?

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, $x = \frac{- 10 \pm \sqrt{64}}{2} = - 5 \pm \frac{8}{2}$ = -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 $6 x^{2} + x - 2 = 0$ as x = $- \frac{2}{3}$ and x = $\frac{1}{2}$
- "Reverse" the method above to factorise
  - $6 x^{2} + x - 2 \equiv (3 x + 2) (2 x - 1)$
- Warning: a calculator gives solutions to 12x² + 2x – 4 = 0 as x = $- \frac{2}{3}$ and x = $\frac{1}{2}$
  - But 12x² + 2x – 4 $≢ (3 x + 2) (2 x - 1)$
  - the right-hand side expands to 6x² + ... ,not 12x² + ...
  - Correct this by multiplying the right by 2
  - 12x² + 2x – 4 $\equiv 2 (3 x + 2) (2 x - 1)$

Exam Tip

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 $x^{2} - 7 x + 2 = 0$ , 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

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

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

$x = \frac{7 \pm \sqrt{41}}{2}$

This is as simplified as possible

$x = \frac{7 \pm \sqrt{41}}{2}$

(b)

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

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

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

Use a calculator to find each solution

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

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

$(2 x - 9) (8 x - 5) = 0$

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 $x^{2} + 6 x + 5 = 0$

This question wants you to complete the square first
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 $2^{x}$

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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