A quadratic equation contains only up to and including $$x^2$$. There are many ways to solve quadratics. All quadratic equations can be written in the form $$ax^2 + bx + c = 0$$ where $$a$$, $$b$$ and $$c$$ are numbers ($$a$$ cannot be equal to 0, but $$b$$ and $$c$$ can be).

Here are some examples of quadratic equations.

• $$2x^2 - 2x - 3 = 0. a = 2, b = -2$$ and $$c = -3$$
• $$2x(x+3) = 0$$. This will expand to $$2x^2 + 6x = 0. a = 2, b = 6$$ and $$c = 0$$
• $$(2x + 1)(x - 5) = 0$$. This will expand to $$2x^2 - 9x - 5 = 0. a = 2, b = -9$$ and $$c = -5$$
• $$x^2 + 2 = 4$$. This will rearrange to $$x^2 - 2 = 0. a = 1, b = 0$$ and $$c = -2$$
• $$3x^2 = 48$$. This will rearrange to $$3x^2 – 48 = 0$$. $$a = 3, b = 0$$ and $$c = -48$$ (in this example $$c = -48$$, but has been rearranged to the other side of the equation)
• $$3x^2 = 48$$ is an example of a quadratic equation that can be solved simply.

Divide both sides by 3

## Solving by factorising

If the of two numbers is zero then one or both of the numbers must also be equal to zero. Therefore if $$ab = 0$$, then $$a = 0$$ and/or $$b = 0$$.

If $$(x + 1)(x + 2) = 0$$, then $$x + 1 = 0$$ or $$x + 2 = 0$$, or both. Factorising quadratics will also be used to solve the equation.

Expanding the brackets $$(x + 2)(x + 3)$$ gives $$x^2 + 3x + 2x + 6$$, which simplifies to $$x^2 + 5x + 6$$. is the reverse process of expanding brackets, so factorising $$x^2 + 5x + 6$$ gives $$(x + 2)(x + 3)$$.

### Example

Solve $$x(x + 3) = 0$$.

The product of $$x$$ and $$x + 3$$ is 0, so $$x = 0$$ or $$x + 3 = 0$$, or both.

$\begin{array}{rcl} x + 3 & = & 0 \\ -3 && -3 \\ x & = & -3 \end{array}$

$$x = 0$$ or $$x = -3$$.

Question

Solve $$(x + 1)(x - 5) = 0$$.

The product of $$x + 1$$ and $$x - 5$$ is 0, so $$x + 1 = 0$$ or $$x – 5 = 0$$.

$\begin{array}{rcl} x + 1 & = & 0 \\ -1 && -1 \\ x & = & -1 \end{array}$

$\begin{array}{rcl} x - 5 & = & 0 \\ + 5 && + 5 \\ x & = & 5 \end{array}$

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

Question

Solve $$x^2 + 7x + 12 = 0$$.

The quadratic will be in the form $$(x + a)(x + b) = 0$$.

Find two numbers with a product of 12 and a sum of 7.

$$3 \times 4 = 12$$, and $$3 + 4 = 7$$, so $$a$$ and $$b$$ are equal to 3 and 4. This gives:

$(x + 3)(x + 4) = 0$

The product of $$x + 3$$ and $$x + 4$$ is 0, so $$x + 3 = 0$$ or $$x + 4 = 0$$, or both.

$\begin{array}{rcl} x + 3 & = & 0 \\ - 3 && - 3 \\ x & = & -3 \end{array}$

$\begin{array}{rcl} x + 4 & = & 0 \\ - 4 && - 4 \\ x & = & -4 \end{array}$

$$x = -3$$ or $$x = -4$$