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## Algebra (all content)

### Course: Algebra (all content) > Unit 9

Lesson 7: The quadratic formula- The quadratic formula
- Understanding the quadratic formula
- Using the quadratic formula
- Worked example: quadratic formula
- Worked example: quadratic formula (example 2)
- Worked example: quadratic formula (negative coefficients)
- Quadratic formula
- Using the quadratic formula: number of solutions
- Number of solutions of quadratic equations
- Proof of the quadratic formula
- Quadratic formula review
- Discriminant review
- Quadratic formula proof review

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# Worked example: quadratic formula (example 2)

Sal solves the equation -x^2+8x=1 by first bringing it to standard form and then using the quadratic formula. Created by Sal Khan and Monterey Institute for Technology and Education.

## Want to join the conversation?

- At3:14, why can't you just simplify the square root of 64-4 into 8-2. Why does that change your answer so much?(12 votes)
- You cannot separate square roots into parts if they have addition/subtraction between terms inside.(31 votes)

- For people with question about how ax^2+bx+c=0 turned into x=-b +/- The square root of b^2-4ac all divided by 2a then here is the steps.

1.General Form.

Ax^2+bx+c=0

2.Subtract C

ax^2+bx=-c

3.Divide a

ax^2+bx all divided by a=-c/a

4.Middle term

1/2*b/a=(b/2a)^2= b^2/4a^2

5.Add b^2/4a^2

x^2+b/a*X+b^2/4a^2=-c/a+b^2/4a^2

6. Factor

(x+b/2a)^2 = -c/a+b^2/4a^2

7. Simplify

-C/a+b^2/4a^2 = -c/a*4a/4a+b^2/4a^2 = -4ac/4a^2+b^2/4a^2

8.New equation

(X+b/2a)^2 = b^2-4ac all divided by 4a^2

9.Square root

Square root of (x+b/2a)^2 = square root of b^2-4ac/4a^2

10. New equation

x+b/2a= +/- Square root of b^2-4ac/4a^2

11.Simplify

x+b/2a= +/- Square root of b^2-4ac all divided by 2a

12.Subtract b/2a

x = +/- Square root of b^2-4ac all divided by 2a - b/2a

13. The Answer

x=-b +/- square root of b^2-4ac all divided by 2a.

Hope this is helpful for the curious ones.(14 votes) - 0:34Does the equation HAVE to equal zero?(9 votes)
- In the quadratic form, ax^2+bx+c=0, and cubic equation, ax^3+bx^2+cx+d=0. And, it depends what formula that you need to use.(1 vote)

- This may seem like a stupid question, -50=10t-5t^2 .... I'm stuck(4 votes)
- In this example, you can start by trying to find if there is some common divisor of all the elements of your equation, or other way to get t^2 without the 5.(7 votes)

- How would I find for x in this equation (x-6)(x+9)=0. I mean what is the formula to find this type of particular equation.(1 vote)
- Take each thing in the parenthesis and set it equal to 0. Then solve for x in each.

x - 6 = 0

x = 6

x + 9 = 0

x = -9

So x equals -9 or 6.(10 votes)

- Why do we say x is equal to something or/ and x is equal to something, but shouldn't it just be and because, we need both values of x in order to draw a parabola, which is a quadratic?(4 votes)
- Think of it this way: Can anything assume two values at the same time? This is not possible, as far as I know. By saying
`x`

is equal to this`and`

that, you are essentially saying that`x`

*can*equal two things at once, which is not the case. So, I always use`or`

. But yes, you do need both x values to draw a parabola.(4 votes)

- Does it have to be -8+2√15/-2 and -8-2√15/-2? or can it be -8+60/-2 and -8-60/-2? and Why?(5 votes)
- At3:02why does Sal cancel out the -1 cancel out as. When they are multiplied together that will give +1 and then we have to add that to the equation. Please help(3 votes)
- I think you are saying the same thing as Sal is, when he says they "cancel out" he is implying they equal to one, but multiplying by 1 does not do anything for you, so you just get -4. So he cancels both the two negatives and multiplying by 1 in one step, you do not add 1 to the equation, you do -4(1).

The way I try to teach my students is to determine the sign, then do the math, so 3 negative signs gives a negative answer and 4*1*1 =4, so you get -4 which is the same.(2 votes)

- At4:46, could've I just put ± instead of ∓? If I did, would've I gotten it wrong? When should I use which one?(2 votes)
- Sal noted that it is really the same thing, but the correct symbol is ±, so you would be correct in doing that. Sal uses what it would actually do to make a point.(3 votes)

- Sometimes Sal refers to the
*square root symbol*as the**square root sign**or the**radical**, and I'm wondering which word is better for it.

Also, why doesn't Sal divide √15 by that -2?(2 votes)- 1) The symbol √ is called a radical. If you are working with square roots saying it is a square root is more precise than saying radical. The reason is that there are multiple types of radicals: square roots, cube roots, 4th roots, etc.

2) When we reduce fractions, we divide out commone factors. In Sal's fraction, the numberator has 2 factors: 2 and √15. Once the 2 divided out with the -2 in the denominator, the denomiantor becomes 1. So, there is no additional factor to divide with the √15. Even if you had √15/2, there is no common factor to cancel out.

Hope this helps.(2 votes)

## Video transcript

Use the quadratic formula to
solve the equation, negative x squared plus 8x is equal to 1. Now, in order to really use the
quadratic equation, or to figure out what our a's, b's and
c's are, we have to have our equation in the form, ax
squared plus bx plus c is equal to 0. And then, if we know our a's,
b's, and c's, we will say that the solutions to this equation
are x is equal to negative b plus or minus the square root
of b squared minus 4ac-- all of that over 2a. So the first thing we have to
do for this equation right here is to put it
in this form. And on one side of this
equation, we have a negative x squared plus 8x, so that looks
like the first two terms. But our constant is on
the other side. So let's get the constant on the
left hand side and get a 0 here on the right hand side. So let's subtract 1 from both
sides of this equation. The left hand side of the
equation will become negative x squared plus 8x minus 1. And then the right hand
side, 1 minus 1 is 0. Now we have it in that form. We have ax squared
a is negative 1. So let me write this down.
a is equal to negative 1. a is equal to negative 1. It's implicit there, you could
put a 1 here if you like. A negative 1. Negative x squared is the same
thing as negative 1x squared. b is equal to 8. So b is equal to 8, that's
the 8 right there. And c is equal to negative 1. That's the negative
1 right there. So now we can just apply
the quadratic formula. The solutions to this equation
are x is equal to negative b. Plus or minus the square root
of b squared, of 8 squared, minus 4ac-- let me do it in that
green color --minus 4, the green is the part
of the formula. The colored parts are the things
that we're substituting into the formula. Minus 4 times a, which is
negative 1, times negative 1, times c, which is
also negative 1. And then all of that-- let me
extend the square root sign a little bit further --all
of that is going to be over 2 times a. In this case a is negative 1. So let's simplify this. So this becomes negative 8, this
is negative 8, plus or minus the square root
of 8 squared is 64. And then you have a negative 1
times a negative 1, these just cancel out just to be a 1. So it's 64 minus is 4. That's just that 4 over there. All of that over negative 2. So this is equal to negative
8 plus or minus the square root of 60. All of that over negative 2. And let's see if we can
simplify the radical expression here, the
square root of 60. Let's see, 60 is equal
to 2 times 30. 30 is equal to 2 times 15. And then 15 is 3 times 5. So we do have a perfect
square here. We do have a 2 times
2 in there. It is 2 times 2 times
15, or 4 times 15. So we could write, the square
root of 60 is equal to the square root of 4 times the
square root of 15, right? The square root of 4 times
the square root of 15, that's what 60 is. 4 times 15. And so this is equal to-- square
root of 4 is 2 times the square of 15. So we can rewrite this
expression, right here, as being equal to negative 8 plus
or minus 2 times the square root of 15, all of that
over negative 2. Now both of these terms right
here are divisible by either 2 or negative 2. So let's divide it. So we have negative 8 divided
by negative 2, which is positive 4. So let me write it over here. Negative 8 divided by negative
2 is positive 4. And then you have this
weird thing. Plus or minus 2 divided
by negative 2. And really what we have
here is 2 expressions. But if we're plus 2 and we
divide by negative 2, it will be negative 1. And if we take negative 2 and
divide by negative 2, we're going to have positive 1. So instead of plus or minus, you
could imagine it is going to be minus or plus. But it's really the same thing. Right? It's really now minus or plus. If it was plus, it's now
going to be a minus. If it was a minus, it's now
going to be a plus. Minus or plus 2 times the
square root of 15. Or another way to view it is
that the two solutions here are 4 minus two roots of 15,
and 4 plus two roots of 15. These are both values of x
that'll satisfy this equation. And if this confuses you, what I
did, turning a plus or minus into minus plus. Let me just take a little
bit of an aside there. I could write this expression
up here as two expressions. That's what the plus
or minus really is. There's a negative 8 plus 2
roots of 15 over negative 2. And then there's a negative
8 minus 2 roots of 15 over negative 2. This one simplifies to--
negative 8 divided by negative 2 is 4. 2 divided by negative
2 is negative 1. 2 times a 4 minus the
square root of 15. And then over here you have
negative 8 divided by negative 2, which is 4. And then negative 2 divided by
negative 2, which is plus the square of 15. And I just realized I made
a mistake up here. When we're dividing a 2 divided
by negative 2, we don't have this 2 over here. This is just a plus or
minus the root of 15. We just saw that when
I did it out here. So this is minus the
square root of 15. And this is plus the
square root of 15. So the two solutions for this
equation-- It's good that I took that little hiatus there,
that little aside there. The two solutions could be 4
minus the square root of 15, or x, or and, x could be 4 plus
the square root of 15. Either of those values of x
will satisfy this original quadratic equation.