Quadratic Equations

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Notes

Solving Quadratics by Factorising

Rearrange into **ax² $+ bx + c =$ 0** with zero on one side.
Factorise the quadratic and set each bracket equal to zero.
If $(x + 4)(x - 1) = 0$ , then $x + 4 = 0$ or $x - 1 = 0$ .
For brackets with coefficients, e.g. $(2x - 3)(3x + 5) = 0$ , solve $2x - 3 = 0$ and $3x + 5 = 0$ .
If x is a factor, e.g. $x(x - 4) = 0$ , solutions are $x = 0$ and $x = 4$ .
Do **not** divide both sides by x; you will lose a solution.
Use a calculator to check factorisation: if solutions are integers or fractions, the quadratic factorises.

The Quadratic Formula

Formula: **x $= (-b \pm \sqrt{b^{2} - 4ac}) /$ (2a)** for $ax^{2} + bx + c = 0 (a \ne 0)$ .
Read off a, b, c and substitute carefully, using brackets for negative numbers.
Simplify using a calculator or by hand; round as required (e.g. 2 d.p., 3 s.f.).
The **discriminant** is $b^{2} - 4ac$ : $if > 0$ two solutions, $= 0$ one solution, $< 0$ no real solutions.
If the discriminant is a perfect square, the quadratic factorises with integers.
Always show working; calculators can be used to check answers.

Completing the Square

Rewrite **x² + bx** as **(x $+ p)^{2} -$ p²** where $p = \frac{b}{2}$ .
For $x^{2} + bx + c$ , complete square: $(x + p)^{2} - p^{2} + c$ , then simplify numbers.
If coefficient $a \ne 1$ , factorise a out of $x^{2}$ and x terms first: $a[x^{2} + (\frac{b}{a})x] + c$ .
Then complete square inside brackets and multiply through by a.
The turning point of $y = (x + p)^{2} + q$ is at **(-p, q)**; for $y = a(x + p)^{2} + q$ , same coordinates.
Turning point is minimum if $a > 0$ , maximum if $a < 0$ .
Check your answer by expanding the completed square.

Solving by Completing the Square

To solve $x^{2} + bx + c = 0$ , complete square: $(x + p)^{2} - p^{2} + c = 0$ .
Rearrange to **(x $+ p)^{2} = p^{2} -$ c**, then take square roots: $x + p = \pm \sqrt{p^{2} - c}$ .
Solve for $x: x = -p \pm \sqrt{p^{2} - c}$ .
If $a \ne 1$ , divide both sides by a first (only for solving, not rewriting).
Answers are often in exact (surd) form.
Do **not** expand the squared bracket back out when solving.

Deciding the Quadratic Method

Use **factorisation** when the question says 'solve by factorising' or for simple quadratics.
Use **quadratic formula** when answers need a given accuracy (e.g. 2 d.p.) or when factorisation is hard.
Use **completing the square** when part (a) asks to complete the square and part (b) uses it to solve.
Completing the square also helps rearrange formulae with $x^{2}$ and x terms.
If in doubt, the quadratic formula always works.
Check solutions with a calculator: integer/fraction solutions mean factorisation works.

Quadratic Formula Diagram

Completing the Square Steps

Factorisation Example

Solving by Completing the Square

Practice questions

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1.What is the standard form of a quadratic equation?
Easy
- A $ax + bx + c = 0$
- B $ax^{2} + bx + c = 0$
- C $ax^{2} + bx + c = y$
- D $ax + b = 0$
2.Solve $x^{2} - 4x = 0$ by factorising.
Easy
- A $x = 0$ or $x = 4$
- B $x = 0$ or $x = -4$
- C $x = 2$ or $x = -2$
- D $x = 4$ only
3.Which of the following is the quadratic formula?
Easy
- A $x = (-b \pm \sqrt{b^{2} - 4ac}) / (2a)$
- B $x = (-b \pm \sqrt{b^{2} + 4ac}) / (2a)$
- C $x = (b \pm \sqrt{b^{2} - 4ac}) / (2a)$
- D $x = (-b \pm \sqrt{b^{2} - 4ac}) / a$
4.Use the quadratic formula to solve $3x^{2} + 7x - 11 = 0$ . Give your answers correct to 2 decimal places.
Medium
- A $x = 1.13$ or $x = -3.46$
- B $x = 1.13$ or $x = 3.46$
- C $x = -1.13$ or $x = 3.46$
- D $x = -1.13$ or $x = -3.46$
5.By completing the square, solve $x^{2} + 6x + 5 = 0$ .
Medium
- A $x = -1$ or $x = -5$
- B $x = 1$ or $x = 5$
- C $x = -1$ or $x = 5$
- D $x = 1$ or $x = -5$
6.Write $x^{2} - 4x + 7$ in the form $(x - a)^{2} + b$ .
Medium
- A $(x - 2)^{2} + 3$
- B $(x - 2)^{2} + 7$
- C $(x - 4)^{2} + 7$
- D $(x - 2)^{2} + 11$
7.The discriminant of $2x^{2} - 3x + 1 = 0$ is:
Medium
- A1
- B9
- C-1
- D17
8.Solve $10m^{2} + 9m - 162 = 0$ .
Hard
- A $m = 3.6$ or $m = -4.5$
- B $m = -3.6$ or $m = 4.5$
- C $m = 3.6$ or $m = 4.5$
- D $m = -3.6$ or $m = -4.5$

Summary

## Quadratic Equations - Revision Notes ### 1. Solving by Factorising - Rearrange to **ax² + bx + c = 0**. - Factorise: if (x + p)(x + q) = 0, then x = -p or x = -q. - For coefficients: (2x - 3)(3x + 5) = 0 ⇒ 2x - 3 = 0 or 3x + 5 = 0 ⇒ x = 3/2 or x = -5/3. - If x is a factor: x(x - 4) = 0 ⇒ x = 0 or x = 4. **Do not divide by x**. ### 2. Quadratic Formula - For ax² + bx + c = 0 (a ≠ 0): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] - **Discriminant** Δ = b² - 4ac: - Δ > 0: two distinct real solutions. - Δ = 0: one repeated solution. - Δ < 0: no real solutions. - Use when answers required to a given accuracy (e.g., 2 d.p.). ### 3. Completing the Square - Rewrite x² + bx as (x + p)² - p², where p = b/2. - For ax² + bx + c: factor a out of first two terms, complete square, simplify. - **Turning point**: y = a(x + p)² + q has vertex at (-p, q). - To solve: (x + p)² = q ⇒ x = -p ± √q. ### 4. Choosing a Method - **Factorising**: quick if factorable; use for simple equations. - **Quadratic formula**: always works; use for decimals/surds. - **Completing the square**: when directed or for deriving turning point. ### Key Points - Always ensure equation = 0 before solving. - Check calculator solutions to verify factorisation. - For exact surd form, simplify √(b² - 4ac) fully.

Key terms

Quadratic equation: An equation of the form ax² + bx + c = 0, where a ≠ 0.
Factorisation: Writing a quadratic as a product of two linear factors, e.g., (x + p)(x + q) = 0.
Quadratic formula: x = (-b ± √(b² - 4ac)) / (2a), used to solve any quadratic equation.
Discriminant: The value b² - 4ac under the square root in the quadratic formula; determines number of real solutions.
Completing the square: Rewriting ax² + bx + c as a(x + p)² + q to find turning point or solve.
Turning point: The vertex of a quadratic graph; minimum if a > 0, maximum if a < 0.

Slides

Solving Quadratics by Factorising

Rearrange the equation so that one side is zero: $ax^{2} + bx + c = 0$ .
Factorise the quadratic into two brackets.
Set each bracket equal to zero and solve for x.
If x is a factor, one solution is $x = 0$ .

Solve $(x - 3)(x + 7) = 0$ . What are the solutions?

$x = 3$ or $x = -7$
$x = -3$ or $x = 7$
$x = 3$ or $x = 7$
$x = -3$ or $x = -7$

The Quadratic Formula

For $ax^{2} + bx + c = 0$ , the quadratic formula is $x = [-b \pm \sqrt{b^{2} - 4ac}] / (2a)$ .
Substitute a, b, and c carefully, using brackets for negative numbers.
The discriminant $b^{2} - 4ac$ determines the number of solutions.
Use the formula when the question asks for decimal or exact surd answers.

The quadratic formula is $x = [-b \pm \sqrt{b^{2} - 4____}] / (2a)$ .

Completing the Square

Rewrite $x^{2} + bx$ as $(x + p)^{2} - p^{2}$ , where $p = \frac{b}{2}$ .
For a quadratic $x^{2} + bx + c$ , complete the square to get $(x + p)^{2} + q$ .
If there is a coefficient a in front of $x^{2}$ , factor it out of the first two terms first.
The turning point of $y = (x + p)^{2} + q$ is at (-p, q).

Complete the square for $x^{2} + 6x + 5$ . What is the result?

$(x + 3)^{2} - 4$
$(x + 3)^{2} + 4$
$(x + 6)^{2} - 31$
$(x + 3)^{2} - 14$

Solving by Completing the Square

To solve $x^{2} + bx + c = 0$ , complete the square to get $(x + p)^{2} = p^{2} - c$ .
Then take square roots: $x + p = \pm \sqrt{p^{2} - c}$ .
Finally solve for $x: x = -p \pm \sqrt{p^{2} - c}$ .
If there is a coefficient a, divide both sides by a first (only for solving).

Solve $(x + 3)^{2} = 16$ by taking square roots: $x + 3 = \pm ____$ . Then $x = -3 \pm 4$ , so $x = 1$ or $x = -7$ .

Deciding the Quadratic Method

Use factorisation when the quadratic factorises easily or when asked.
Use the quadratic formula when decimal or significant figure accuracy is required.
Use completing the square when the question directs you to, or to find turning points.
If in doubt, the quadratic formula always works.

Which of these are true?

Factorisation is always the fastest method.
The quadratic formula can be used for any quadratic equation.
Completing the square can help find the turning point.
If the discriminant is negative, there are two real solutions.

Worked Example: Factorising

Solve $x^{2} + 3x - 10 = 0$ by factorising.
Factorise: $(x - 2)(x + 5) = 0$ .
Set each bracket to zero: $x - 2 = 0$ → $x =$ 2; $x + 5 = 0$ → $x = -5$ .
Solutions: $x = 2$ or $x = -5$ .

Solve $5x^{2} - x = 0$ by factorising. What are the solutions?

$x = 0$ or $x = \frac{1}{5}$
$x = 0$ or $x = 5$
$x = \frac{1}{5}$ only
$x = 0$ only

Worked Example: Quadratic Formula

Solve $3x^{2} - 2x - 4 = 0$ using the quadratic formula.
$a = 3, b = -2, c = -4$ .
$x = [2 \pm \sqrt{4 + 48}] / 6 = [2 \pm \sqrt{52}] / 6 = [2 \pm 2\sqrt{13}] / 6 = (1 \pm \sqrt{13})/3$ .
Exact solutions: $x = (1 + \sqrt{13})/3$ or $x = (1 - \sqrt{13})/3$ .

For $3x^{2} - 2x - 4 = 0$ , the discriminant $b^{2} - 4ac = (-2)^{2} - 4(3)(-4) = 4 + ____ = 52$ .

Worked Example: Completing the Square

Solve $2x^{2} - 8x - 24 = 0$ by completing the square.
Divide by $2: x^{2} - 4x - 12 = 0$ .
Complete square: $(x - 2)^{2} - 4 - 12 = 0$ → $(x - 2)^{2} = 16$ .
Take square roots: $x - 2 = \pm 4$ → $x = 6$ or $x = -2$ .

Solve $x^{2} + 6x + 5 = 0$ by completing the square. What are the solutions?

$x = -1$ or $x = -5$
$x = 1$ or $x = 5$
$x = -1$ or $x = 5$
$x = 1$ or $x = -5$

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