Use the discriminant

to state the number and type of solutions

for the equation negative 3x squared plus

5x minus 4 is equal to 0. And so just as a

reminder, you’re probably wondering what

is the discriminant. And we can just

review it by looking at the quadratic formula. So if I have a quadratic

equation in standard form, ax squared plus bx

plus c is equal to 0, we know that the quadratic

formula, which is really just derived from completing

the square right over here, tells us that the roots

of this, or the solutions of this quadratic

equation are going to be x is equal to negative b

plus or minus the square root of b squared minus 4ac,

all of that over 2a. Now, you might know

from experience applying this little bit,

we’re going to get different types of

solutions depending on what happens under the

radical sign over here. As you could imagine, if

what’s under the radical sign over here is

positive, then we’re going to get an

actual real number as its principal square root. And when we take the positive

and negative version of it, we’re going to get

two real solutions. So if b squared minus

4ac– and this is what the discriminant

really is, it’s just this expression

under the radical sign of the quadratic formula. If this is greater

than 0, then we’re going to have two real

roots or two real solutions to this equation right here. If b squared minus

4ac is equal to 0, then this whole thing is

just going to be equal to 0. It’s going to be the plus or

minus square root of 0, which is just 0. So it’s plus or minus 0. Well, when you

add or subtract 0, that doesn’t change

the solution. So the only solution is going

to be negative b over 2a. So you’re only going to

have one real solution. So this is going to be 1– I’ll

just write the number 1– 1 real solution. Or you could say you have

a repeated root here. You could say you’re

having it twice. Or you could say one real

solution, or one real root. Now, if b squared minus

4ac were negative, you might already

imagine what will happen. If this expression right

over here is negative, we’re taking the square

root of a negative number. So we would then get an

imaginary number right over here. And so we would add or subtract

the same imaginary number. So we’ll have two

complex solutions. And not only will we have

two complex solutions, but they will be the

conjugates of each other. So if you have one

complex solution for a quadratic equation,

the other solution will also be a complex solution. And it will be its

complex conjugate. So here we would have

two complex solutions. So numbers that have a real

part and an imaginary part. And not only are

they just complex, but they are the

conjugates of each other. The imaginary parts

have different signs. So let’s look at b squared

minus 4ac over here. This is our a, this is

our b, and this is our c. Let me label them. a, b, c. And I can do that, because we’ve

written it in standard form. Everything is on one side. Or in particular,

the left hand side. We have a 0 on the

right hand side. We’ve written it

in descending, I guess, power form, or

the descending degree. Or we have our second

degree term first, then our first degree term,

then our constant term. And so we can evaluate

the discriminant. b is 5. So b squared is 5 squared, minus

4 times a, which is negative 3, times c, which is negative

4– I have to be careful. c is this whole thing.

c is negative 4. And so I don’t know if I said

4 earlier, but c is negative 4. We have to make sure we take

the sign into consideration. So times c, which is

negative 4 over here. And so this is 25. And then negative

3 times negative 4, that is positive 12. And then 4 times 12 is 48. But we have a negative out here. So 25 minus 48. And 25 minus 48, we don’t

even have to do the math. We can just say that

this is definitely going to be less than 0. You can actually figure it out. This is equal to negative

13, if I did– oh no, sorry, negative 23, which

is clearly less than 0. So our discriminant in this

situation is less than 0. So we are going to have

two complex roots here, and they’re going to be

each other’s conjugates.