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Discriminant for types of solutions for a quadratic | Algebra II | Khan Academy


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.

Bernard Jenkins

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