# Quadratic Equations

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## Quadratic Equations

**Quadratic equations are equations that can be written:**

**ax ^{2} +bx + c = 0**

where **a, b** and **c** are constants (numbers) and 'a' cannot be zero (-there must be some x^{2} 's!) You might also see them written in the form (x + p)(x + q) = 0 but we?ll come to that later.

**You solve them by finding the value(s) of x which make them equal to zero. There are several methods:**

**1. Factorising**.

**2. The quadratic formula.**

**3. Completing the square** (most exam boards leave this to AS level).

**4. Using the graph. **

This means **'putting in brackets'.** (You also need to be able to multiply the brackets out to get the equation back to the form: ax^{2} + bx + c = 0).

You need to find a number to go in each bracket. **Use the following conditions to find the two numbers you need:**

**1.** They must **multiply** to give 'c'.

**2.** They must **add** to give 'b'.

**3.** If 'c' is positive both numbers must have the same sign (both positive or both negative). If 'c' is negative then the numbers must be of opposite sign.

There's no magic solution. You just need to **practise, practise, practise!**

Then, as the brackets multiply to give zero **one of the brackets must be zero!**

**This gives you your solution(s).** (Quadratics will either have 0, 1 or 2 solutions).

**Here's an example:**

**Solve x ^{2} + 3x - 10 = 0.**

It's already written in the form: **ax ^{2} + bx + c = 0,** so we don't need to rearrange anything. We need two numbers that multiply to give -10 (which means they're of opposite sign) and add to give 3.

There are 4 possibilities: -1 and 10, 1 and -10, -2 and 5, 2 and -5.

**The only pair that adds to give +3 is -2 and 5 giving:**

**(x -2)(x + 5) = 0**

If (x - 2) = 0 we get the solution x = 2

If (x + 5) = 0 we get the solution x = -5

**Try solving x ^{2} + 5x +6 = 0:**

If an equation won't factorise (the solutions might be decimals or fractions) then you can try the formula. It looks really complicated but once you've practised using it, it will seem much easier (honest!). Anyway, it's based on the form: **ax ^{2} + bx + c = 0, and here it is:**

You just substitute the numbers in and hey presto! One solution comes from using the + sign in front of the square root and the other solution by using the - sign.

**Here's an example:**

**Solve x ^{2} - 2x - 4 = 0 using the formula.**

Here a = **1**, b = -**2**, and c = -**4**. So we can put these values into the formula:

**Simplifying:**

Putting this **carefully **into your calculator (see 'Calculator' section) gives the solutions:

x = 3.24, and x = -1.24 (both to 2 decimal places).

**Heres a worked example, follow it through:**

Well, the important bit of the formula is the bit inside the square root: **b ^{2}- 4ac**

**If b ^{2}- 4ac is positive**, there are

**two**solutions.

**If b ^{2}- 4ac is zero,** there is

**one**solution.

**If b ^{2}- 4ac is negative,** there are

**no**solutions.

To look at graphs of linear and quadratic equations, go to Graphs Learn-its.