[SOUND] Let's learn about the discriminant.

[SOUND] Let's find the number of real solutions of each of these 3 quadratic

equations. [SOUND] Let's first recall the quadratic

formula. It gives us the solutions to quadratic

equations of this form ax ^ 2 + bx + c = 0, and those solutions are given by x =

-b + or - the square root of b ^ 2 - 4ac all / 2a.

However, here we're not asked to find the solutions, but rather the number of real

solutions. And this quantity here under this square

root is what determines that. This quantity is what we call the

discriminate. And if we think of it, this discriminant,

this B squared minus 4ac, is a positive real number, then there be 2 real

solutions to this quadratic equation. Negative b plus the square root of that

positive number divided by 2a, and negative b minus the square root of that

positive number divided by 2a. So in this case there are two real

solutions. Now when the discriminant is identically

equal to 0, there is only one real solution.

Namely, x equal negative b divided by 2a, because in the numerator we'd be adding

and subtracting 0. And finally, what if that discriminant is

a negative number? Then, there are no real solutions, because a square root of

a negative number does not yield any real number.

We'd have complex solutions but not real solutions.

So let's compute this discriminant for each of our 3 equations here.

So for the 1st equation, we have a is equal to 3, b is equal to negative 6 and

b is equal to 5. So, therefore, our discriminant, b

squared minus 4ac is equal negative 6 squared minus 4 times 3 times 5, which is

equal to 36 minus 60, which is equal to negative 24, which is a negative number.

Which means we're down in this third case, aren't we? Which means our first

equation has no real solutions. Alright, what about our second equation

here? We have that A is equal to 3, B is equal

to negative 6, and c is equal to 1.

Which means our discriminant then, this b squared minus 4ac is equal to negative 6

squared minus 4 times 3 times 1, which is equal to, 36 minus 12.

Which is equal to 24, which is greater than zero, which means we're in this

first case down here. And therefore, our second equation has

two real solutions. Alright, and what about our last equation

here? Our a is equal to 3, b is equal to negative 6 and c is equal to 3, which

means that the discriminant b squared minus 4ac is equal to negative 6 squared

minus 4 times 3, times 3, which is is equal to 36 minus 36, which is equal to

0. And therefore, we're in this second case

down here, which means our third equation has one

real solution. So the discriminant helps us determine

how many real solutions we have to a given quadratic equation.

Thank you and we'll see you next time. [SOUND]