If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Using the quadratic formula: number of solutions

Sal determines how many solutions the equation x²+14x+49=0 has by considering its quadratic formula, and more specifically, its discriminant. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

• So when we say we don't have "real solutions" we mean that there is no point were the parabola is touching on the x axis which is the R or real number line? And when we say we have one solution which is observable from the discriminant we say that there is only one point that touches the x axes (Real number line) and therefore it has to be its vertex? • i was taught that if answer was positive it is called 2 different real numbers, if it was 0 it was called real root, and if it was negative it was imaginary numbers? is this just a different method or is this wrong... •  That's correct. 2 different real numbers = 2 solutions, real root = 1 solution (technically 2 but because both roots are the same there's only one solution) and imaginary numbers = no real solution (because solution is imaginary).
• what real life situations are there where you need the quadratic equation • At why is a=1? • What is the meaning of having two or more answers for one equation or one expression ? and how is it applicable in the real physical world? • That's a very good question. From the mathematical standpoint, two or more answers just means that there is more than one value for the variable that will solve the equation. Thinking of it graphically, if you are solving a linear equation, such as x-3 = 0, this is equivalent to looking at a line y = x-3 and seeing where it crosses the x axis (where y=0). A line can only cross another line in at most one place, so there can be at most one solution (x=3). A parabola, though, curves, so it can cross the x axis in two places. So if you have an equation like x^2 + 5x + 6 = 0, it can have two solutions.
In the real physical world, there are cases in which both these solutions will be valid, and cases where they won't, so you need to pay attention to what the solutions actually mean. For example, if you throw a ball in the air, its height over time can be described as a parabola (quadratic equation). If you are asked to calculate when it hits the ground, you will get two solutions, but both of those can't be right, because physically that makes no sense. Often in that case you might get one solution with positive time, and one solution at a negative value of time, so that one you'd throw out.
However, you might also have a problem in which there are two objects moving (maybe, two vehicles of some kind), and one is moving in a straight line and one is moving in a parabola. You might be asked where their trajectories overlap (not necessarily at the same time). Because a parabola and a line can intersect in two places, you might get two answers, and both might be correct.
I'm sure there are better real world examples! I hope this helps some, though.
• why does b^2-4ac < 0 no real solution when it two imaginary solutions • -8x^2 + 41x - 5 = (by grouping:) (-8x+1)(x-5)
When I multiply the first 2 factors I get the original equation again.
Trying this with the second set of factors, i end up with opposite signs (8x^2 - 41x + 5).
What am I doing wrong ? • If you expand (x-1/8)(x-5) you actually end up with x^2-(41/8)x+5/8 and not
8x^2 - 41x + 5. This makes sense because the quadratic formula will give you the roots to a quadratic equation, but remember, there are infinitely many quadratics with these roots. Think of it this way: you start with the quadratic equation: -8x^2 + 41x - 5=0 but you can do anything to rearrange this equation as long as you do it to both sides (and do not multiply by zero since that would not help at all). So if you wanted to divide both sides by 2, you would get:
-4x^2+(41/2)x-5/2 = 0 and from this equation you see that -4x^2+(41/2)x-5/2 is yet another quadratic with the same roots as -8x^2 + 41x - 5. You can keep on going and you will see that you will get different-looking quadratics with the same roots. More specifically, you get to (x-1/8)(x-5) by dividing your original quadratic -8x^2 + 41x - 5 by -8 on both sides of the quadratic equation.

(-1/8)(-8x^2 + 41x - 5) = (x-1/8)(x-5)

Hope that helps.
• During these kind of problems my math teacher asks whether each solution is rational and irrational, but how do I know this? • When you solve a quadratic using the quadratic formula, you do a lot of simplification, including simplifying the square root. It the square root goes away because it contained a perfect square, then your answers will be rational numbers. If you still have a radical after simplifying as much as possible, then you have answers that are irrational numbers. • 