Completing the square and the quadratic formula
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Solving Quadratic Equations by Square Roots
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Solving quadratics by taking the square root
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Solving Quadratic Equations by Completing the Square
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Completing the square 1
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Completing the square 2
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How to Use the Quadratic Formula
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Proof of Quadratic Formula
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Quadratic formula
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Example: Complex roots for a quadratic
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Discriminant of Quadratic Equations
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Discriminant for Types of Solutions for a Quadratic
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Solutions to quadratic equations
Discriminant for Types of Solutions for a Quadratic Discriminant for Types of Solutions for a Quadratic
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- Use the discriminant to state the number
- and type of solutions for the equation
- -3*x*²
- -3*x*² + 5*x*
- -3*x*² + 5*x* - 4
- -3*x*² + 5*x* - 4 = 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*²
- *ax*² + *bx*
- *ax*² + *bx* + *c*
- *ax*² + *bx* + *c* = 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* =
- *x* = (-*b*)
- *x* = (-*b* ± √())
- *x* = (-*b* ± √(*b*²))
- *x* = (-*b* ± √(*b*² - 4*ac*))
- all of that over (2*a*).
- Now, you might know from experience
- applying this a little bit,
- we're going to get different types of solutions
- depending on what happens
- under the radical sign over here.
- As you can 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*² - 4*ac*, 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 zero,
- then we're going to have two real roots,
- then we're going to have two real roots,
- or two real solutions to this equation right here.
- If *b*² - 4*ac* = 0,
- then this whole thing is
- just going to be equal to zero,
- so plus or minus the square root of zero,
- (which is just zero)
- so this is plus or minus zero.
- Well, when you add or subtract 0,
- that doesn't change the solution,
- so the only solution is going to be
- -*b* / 2*a*
- So you're only going to have one real solution.
- So this is going to be one--
- --I'll just write the number "1"--
- --one real solution,
- or you could kind of say,
- you have a repeated root here.
- You're kind of having it twice.
- Or you could say one real solution or one real root.
- Now if *b*² - 4*ac* 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.
- So we would add or subtract
- the same imaginary number.
- So we'll have two complex solutions;
- 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 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*² - 4*ac* over here.
- This is our *a*,
- this is our *b*,
- and this is our *c*.
- Let me label them
- Let me label them – *a*
- Let me label them – *a*, *b*
- Let me label them – *a*, *b*, *c*.
- I can do that because we've
- written it in standard form.
- Everything is on one side,
- in particular the left-hand side,
- we have a zero on the right-hand side,
- we've written it in descending power form,
- or descending degree,
- where we have our 2nd degree term first,
- then our 1st degree term,
- then our constant term.
- And so, we can evaluate the discriminant!
- *b* = 5
- *b* = 5, so *b*² = 5²
- 5² - 4
- 5² - 4 • *a*
- 5² - 4 • (-3)
- 5² - 4 • (-3) • c
- 5² - 4 • (-3) • (-4)
- *c* is this whole thing, I have to be careful.
- *c* is *negative* 4, we have to make sure
- we take the sign into consideration,
- so times *c*, which is *negative* 4 over here.
- So this is
- 25 - 4 • (-3) • (-4)
- 25 - 4 • 12
- 25 - 48
- We don't even have to do the math –
- we can just say that this is definitely going to
- be less than zero.
- You can actually figure it out –
- this is equal to negative 23,
- negative 23...
- which is clearly less than zero.
- So our discriminant in this situation is
- less than zero, so we are going to have
- two complex roots here,
- and they are going to be each other's conjugates.
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