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Class 10 (Old)
Course: Class 10 (Old) > Unit 4
Lesson 4: Nature of roots- Using the quadratic formula: number of solutions
- Number of solutions of quadratic equations
- Discriminant for types of solutions for a quadratic
- Equations with equal roots (intermediate)
- Equations with equal roots (advanced)
- Finding nature of roots
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Discriminant for types of solutions for a quadratic
Discriminant for Types of Solutions for a Quadratic. Created by Sal Khan and Monterey Institute for Technology and Education.
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- At, what are conjugates? 2:53(29 votes)
- For any (a+b), its conjugate is (a-b) and vice versa. This is very handy when dealing with roots and complex numbers :)(31 votes)
- Under what circumstance is there only one complex solution?(11 votes)
- There are no circumstances in which you have one and only complex solution.
you either have two real solution or two complex solutions. This is because complex solution occur in pairs for example, a+bi or a-bi.(13 votes)
- The last video (Discriminant of Quadratic Equations) you wrote b² - 4ac < 0 is called NO REAL SOLUTION at, but in this video, why it's called 2 complex solutions? In the same previous video, you wrote b² - 4ac > 0 is called solutions, but you wrote b² - 4ac > 0 is called 2 real roots in this video. What is the difference? 3:45(4 votes)
- By definition complex numbers are "not real", so the two statements are equivalent. If an equation has real solutions it just means that its graph crosses the x axis at some point. If it doesn't have real solutions, the graph of the equation never crosses the x axis. This is because you're solving for x when y equals zero (notice quadratics are always solved equal to zero?), and since there IS NOT an x intercept in an equation that doesn't cross the x axis (sounds obvious I know) you won't get a solution.
There is more confusion about this than there has to be. "Imaginary solutions", "double roots", "conjugate pairs" and all that stuff sounds mysterious, but it only ends up that way because you're trying to find a relationship between a y value of zero, and x. The coordinate for a real number solution would be (x,0), so if there is no x intercept it's game over. But the graph of the equation still exists, you can see it right there on the paper. Sorry if this made no sense, just keep watching the vids, there are plenty on imaginary and complex numbers.(9 votes)
- How do you find the b if the equation is= 7y squared-2=0(2 votes)
- Andrew,
7y² - 2 = 0
And what is the coefficient of y?
I refer to it as being invisible. But you can make it visible by re-writing the equation as
7y² + 0y -2 = 0
So b = 0 when you use the quadratic formula for this equation.(13 votes)
- What would you do if you were trying to solve the discriminant of a quadratic equation with more than one variable and more than 3 terms? (For example x^2-2ax+2a^2+1=0)(6 votes)
- We just do exactly the same thing. You can lump the final two terms together, as neither of them involves x. So here, A = 1, B = -2a and C = 2a^2 + 1 (I've used capital letters for A, B and C since you've already used the variable 'a' in your quadratic). The discriminant is B^2 - 4AC, which is (-2a)^2 - 4(2a^2+1) = 4a^2 - 8a^2 - 4 = -4(a^2 + 1). What does this mean? Well, a^2 is always non-negative (i.e. zero or positive) so the bracketed term a^2+1 is always positive. Thus the discriminant is always -4 times a positive number, so it is always negative. Hence, whatever the value of 'a' in the equation you gave, the discriminant will be negative and the roots will be complex.(3 votes)
- Is there a difference between the roots and the solution?(4 votes)
- They are the same thing, but because distances cannot be negative, sometimes only one root works(1 vote)
- How can you determine if the solution will have two irrational answers?(3 votes)
- How do I know whether there are two complex solutions or one complex solution??(2 votes)
- If the discriminant is negative, then there are always two complex solutions. In fact, if one is a + bi, then then other will always be a - bi.(3 votes)
- in previous video, it said if discriminant (b^2-4ac) < 0 : there is no solutions.
and it says here : two complex solutions ?!(1 vote)- There ARE solutions if the domain extends over all types of numbers. That is, if the domain includes complex numbers, not just real numbers.(5 votes)
- Can someone give examples of Complex Solutions (Including the conjugate)?
I'm imagining solutions of 2+3i and 2-3i or 1+√3i and 1-√3i.. Is this accurate?(2 votes)
Video transcript
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.