How to find how many real solutions an equation has

#b^2-4ac = 4^2 - 4(2)(2) = 0# so this equation has exactly one real solution.

For a quadratic of the form #ax^2+bx+c = 0#
the solution roots are given by the quadratic formula: #x=(-b+-sqrt(b^2-4ac))/(2a)#

The discriminant is the portion #(b^2-4ac)#
if it is negative then the roots include a term which is the square root of a negative number; since there are no Real numbers whose square root is negative, when the discriminant is #<0# there are no Real roots.

If the discriminant is #=0# then #+-sqrt(0)# is a single value (#+0 = -0#)
and the quadratic has exactly one Real solution.

If the discriminant is #>0# then #+-sqrt("discriminant")# represents two different values and the quadratic has two Real solutions.

A quadratic equation is an equation that looks like:

x2 + 4x - 2 = 0.

The general form of this is written as ax2 + 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 x2 + 5x + 6 = 0. Has solutions x = 2 and x = 3.

Case 2: 1 repeated solution - eg x2 + 4x + 4 = 0. Has solution x = 2.

Case 3: No solutions - eg x2 + 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 ax2 + bx + c = 0 as:

x = [-b ± √(b2 - 4ac)]/2a

where the ± signifies that the two solutions are 

x = [-b + √(b2 - 4ac)]/2a and 

x = [-b - √(b2 - 4ac)]/2a.

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 √(b2 - 4ac) is the part that is causing problems! As I said, we cannot take the square root of a negative number, so if b2 - 4ac is negative, we have an error, and no solutions.

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

If b2 - 4ac is positive (>0) then we have 2 solutions.

If b2 - 4ac is 0 then we have only one solution as the formula is reduced to x = [-b ± 0]/2a. So x = -b/2a, giving only one solution.

Lastly, if b2 - 4ac is less than 0 we have no solutions. 

Example:

How many solutions does x2 - 3x + 2 = -1 have?

1) Rearrange to fit the general formula: x2 - 3x + 3 = 0. So a = 1, b = -3 and c = 3.

2) Use the formula: b2 - 4ac = (-3)2 - 4(1)(3) = 9 - 12 = -3.

3) As b2 - 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 b2 - 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!

Video transcript

Determine the number of solutions to the quadratic equation, x squared plus 14x plus 49 is equal to 0. There's a bunch of ways we could do it. We could factor it and just figure out the values of x that satisfy it and just count them. That will be the number of solutions. We could just apply the quadratic formula. But what I want to do here is actually explore the quadratic formula, and think about how we can determine the number of solutions without even maybe necessarily finding them explicitly. So the quadratic formula tells us that if we have an equation of the form ax squared plus bx plus c is equal to 0, that the solutions are going to be-- or the solution if it exists is going to be-- negative b plus or minus the square root of b squared minus 4ac. All of that over 2a. Now the reason why this can be 2 solutions is that we have a plus or minus here. If this b squared minus 4ac is a positive number-- so let's think about this a little bit. If b squared minus 4ac is greater than 0, what's going to happen? Well, then it's a positive number. It's going to have a square root. And then when you add it to negative b you're going to get one value for the numerator, and when you subtract it from negative b you are going to get another value in the numerator. So this is going to lead to two solutions. Now what happens if b squared minus 4ac is equal to 0? If this expression under the radical is equal to 0, you're just going to have the square root of 0. So it's going to be negative b plus or minus 0. And it doesn't matter whether you add or subtract 0, you're going to get the same value. So in that situation, the actual solution of the equation is going to be negative b over 2a. There's not going to be this plus or minus, it's not going to be relevant. You're only going to have one solution. So if b squared minus 4ac is equal to 0, you only have one solution. And then what happens if b squared minus 4ac is less than 0? Well if b squared minus 4ac is less than 0, this is going to be a negative number right here and you're going to have to take the square root of a negative number. And we know, from dealing with real numbers, you can't take the square root. There is no real number squared that becomes a negative number. So in this situation there is no solutions, or no real-- when I say real I literally mean a real number-- no real solution. So let's think about it in the context of this equation right here. And just in case you're curious if whether this expression right here, b squared minus 4ac, has a name, it does. It's called the discriminant. This is the discriminant. That's that part of the quadratic equation. It determines the number of solutions we have. So if we want to figure out the number of solutions for this equation, we don't have to go through the whole quadratic equation, although it's not that much work. We just have to evaluate b squared minus 4ac. So what is b squared minus 4ac? So b is right here, it's 14. So it's 14 squared minus 4 times a, which is 1, times c, which is 49. That c, right there, times 49. What's 14 times 14? Let me do it over here. 14 times 14. 4 times 4 is 16. 4 times 1 is 4. Plus 1 is 56. Put a 0. 1 times 14 is 14. It is 6, 9, 1. It's 196. So this right here is 196. And we can ignore the 1. What's 4 times 49? So 49 times 4. 4 times 9 is 36. 4 times 4 is 16 plus 3 is 190-- or is 19, so you get 196. So this right here is 196. So b squared minus 4ac is 196 minus 196. So 196 minus 196 is equal to 0. So we're dealing with a situation where the discriminant is equal to 0. We only have one solution. And if you want, you could try to find that one solution. This whole part is going to be the square root of 0. It's just going to be 0. So the solution is going to be negative b over 2a. And negative b is-- we could just solve it. Negative b is negative 14 over 2 times a. a is just 1 over 2. So it's equal to negative 7. That's the only solution to this equation. But if you just wanted to know how many solutions, you just have to find out that b squared minus 4ac is 0. So it's only going to have one solution. And there's other ways. You could have actually factored this pretty easily into x plus 7 times x plus 7 and gotten the same result.

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