# How do I know how many solutions a quadratic equation has?

A quadratic equation is an equation that looks like:

x^{2} + 4x - 2 = 0.

The general form of this is written as ax^{2} + bx + c = 0, where a, b and c are all numbers, and x is our unknown variable. In the example above, we would have a = 1, b = 4 and c = -2.

In order to find the number of solutions, we shall split the quadratic equation into 3 cases.

Case 1: 2 unique solutions - eg x^{2} + 5x + 6 = 0. Has solutions x = 2 and x = 3.

Case 2: 1 repeated solution - eg x^{2} + 4x + 4 = 0. Has solution x = 2.

Case 3: No solutions - eg x^{2} + 2x + 4 = 0. Has no solutions.

But how do we know which case we are in? To do this, we take a look at the quadratic formula, which you will hopefully have seen by now. For reference, it gives the solution of the general quadratic ax^{2} + bx + c = 0 as:

*x* = [-*b* ± √(*b*^{2} - 4*ac*)]/2*a*

where the ± signifies that the two solutions are

*x* = [-*b* + √(*b*^{2} - 4*ac*)]/2*a and *

*x* = [-*b* - √(*b*^{2} - 4*ac*)]/2*a.*

In Case 1, this will give two separate answers for x. In Case 2, both answers will be the same.

However, in Case 3 you will likely arrive at an error! This error arises from the fact that we cannot take the square root of a negative number*. This means, that if we are in case 3, then the section √(*b*^{2} - 4*ac*) is the part that is causing problems! As I said, we cannot take the square root of a negative number, so if *b*^{2} - 4*ac *is negative, we have an error, and no solutions.

This is the key to knowing how many solutions we have:

If *b*^{2} - 4*ac *is positive (>0) then we have 2 solutions.

If *b*^{2} - 4*ac *is 0 then we have only one solution as the formula is reduced to *x* = [-*b* ± 0]/2*a. *So *x* = -b/2*a*, giving only one solution.

Lastly, if *b*^{2} - 4*ac *is less than 0 we have no solutions.

**Example:**

How many solutions does x^{2 }- 3x + 2 = -1 have?

1) Rearrange to fit the general formula: x^{2} - 3x + 3 = 0. So a = 1, b = -3 and c = 3.

2) Use the formula: b^{2 }- 4ac = (-3)^{2 }- 4(1)(3) = 9 - 12 = -3.

3) As b^{2 }- 4ac < 0, we have no solutions.

So there you have it! Please get in touch if you require any further assistance.

** ***For those interested/advanced students:** Technically, you CAN take a square root of a negative number. It's beyond the scope of a GCSE course, so if you're confused by anything after this, don't worry! First of all though, I'll explain why nobody has told you this yet.

*Imagine that I asked you to give me the answer to 7 ÷ 3, but you could only use whole numbers. The equation 7 ÷ 3 is equal to 2.33..., but this is not a whole number! So no whole number solutions exist. If I allowed you to use fractions, you could tell me that 7 ÷ 3 is 7/3 or 2 and 1/3.*

*The same idea applies to the problem here. We only have Real numbers (that is, fractions, decimals, whole numbers and "irrational" numbers such as pi) to deal with the question, and if you are asked to take the square root of a negative number, there are no Real solutions! *

*A solution does exist in the "Imaginary" numbers. You don't know about these numbers yet (just like you didn't know about fractions at first). You will learn more about this in A level Further Maths, or perhaps at University, but if this sounds interesting please do check them out via Google.*

*If b ^{2 }- 4ac < 0 then there are no "Real" solutions. *

*However, for your GCSEs, saying that there are no solutions will be good enough for the exam!*