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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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# Proof of the quadratic formula

CCSS.Math: , ,

Sal proves the quadratic formula using the method of completing the square. Created by Sal Khan and CK-12 Foundation.

## Want to join the conversation?

- Why when we took the square root of the right side of the equation, in the numerator we left it with a +/- sign, however with the denominator we were content with making it 2a, and not +/- 2a?(20 votes)
- The plus or minus sign only needs to be on one part of the fraction to be counted

This is because if you have -1/3

And make it (-1)/(-3) it isn't -1/3 anymore it's 1/3 so that isn't correct also (-1)/3 is equal to 1/(-3) as it's still 1/3 and still negative

1/(-3) is more annoying to use and look at so most people do (-1)/3

So that's why he didn't apply the plus or minus on the bottom

Ohhh plus when you derive the equation it's -b/2a and the other value and to stick the two together they need equal denominators so you shove the plus or minus on top not on the bottom that's all of I can think why thanks for asking.(25 votes)

- At5:45, when we take square root of both sides, why does the right side end up with a "plus or minus" designation while the left side stays only positive?(15 votes)
- | Because the equation on the left has implied positive domain because it is represented with a (term)^2 and that represents that any term within the brackets which were positive or negative will result in a positive output. While on the right, we do not yet know the domain of the right so it is correct to assume that there could be a negative root. I'm sorry if i have confused you. :((23 votes)

- At6:00, why did he simply not just take the square root of b squared and leave it as b?(9 votes)
- Because you can't distribute square roots like that. You could if the terms in a radical are products or quotients, but when they are sums and differences(i.e. adding and subtracting), you can't distribute a radical.(16 votes)

- At6:12, why can't you bring the b^2 out from under the radical?

The square root of b^2 is b, right?(7 votes)- B^2 cannot be taken out of the radical even though it is a perfect square. As my teacher tells me, it is "married" to the -4ac, so they cannot be ripped apart like that. Let me show you:

If we have the square root of the quantity 2^2 plus 3^2, you can simplify that into the square root of the quantity 4 plus 9. Then that would equal the square root of 13, which is much different than if you were to take the square root of 2^2, then the same with 3^2. That would mislead you because then your answer would be 5, not the square root of 13. If it were the square root of (b^2 * -4ac), then you could square root the b^2 separately because the square root of x times the square root of y is equal to the square root of (xy).(6 votes)

- At around1:25, Sal divided "a" by everything... but i thought it was illegal to divide by "a" because it might be 0!! (And you can't divide by zero.) So why can Sal do this?(5 votes)
- If a=0 then it's not a quadratic equation any more, "a" has to be more than or equal to 1 in order for it to be a quadratic equation, because the x^2 would dissapear if a=0, so in that way, you can always divide by "a" and there won't be any problems.(9 votes)

- Is there a similar formula for the solutions of a cubic equation?(2 votes)
- Yes, there is a cubic formula as well as a quartic formula. They are not often used because of how difficult they are.

Here is a link to the cubic formula:

http://www.math.vanderbilt.edu/~schectex/courses/cubic/

And if you think that is bad, the quartic formula is truly impressive:

https://upload.wikimedia.org/wikipedia/commons/9/99/Quartic_Formula.svg

There is no general formula possible for a polynomial of 5th degree or higher (at least not one with a finite number of steps). There are methods to find the roots of these polynomials (depending on the details), but not a general formula that always works.(13 votes)

- When Sal is subtracting c/a from b^2/4a^2 if 4a^2 is the common denominator then shouldnt it come out to b^2/4a^2 minus 4a^2c/4a^2? I feel like the numerator in the second term should be squared. 4a^2c. Which would make the answer b^2-4a^2c/4a^2(4 votes)
- He extends the c/a to the common denominator 4a². First of all, what do you have to multiply the denominator a with for it to become the desired 4a²? Well, 4a² is the same as 4*a*a, so if we have a, we need to multiply that by a 4 and another a, so 4a: a * 4a = 4a². And if that's the multiplicatory (not a word, whatever) change we're making in the denominator, we have to multiply the numerator by the same factor in order for the value of the fraction to stay the same. So, c/a = (c*4a)/(a*4a) = 4ac/4a².

Hope that helped!(5 votes)

- What does vertex form of a quadratic equation means? And is this right:

a(x-h)²+k=0

x=h±√-k/a(4 votes)- 1) Well, it is a quadratic equation that can simply find the vertex like the one you have.

2) It is correct.(4 votes)

- how do you find 2 solutions(3 votes)
- You simply solve one solution assuming the square root is positive, and the other assuming the square root is negative.

For example:

In the formula,

3x^2 + 7x + 2 = 0, then

a = 3

b = 7

c = 2

So, using the quadratic formula,

x = (-b +/- (b^2 - 4ac)^ (1/2))/2a

=> x = (-7 +/- (7^2 - 4*3*2)^(1/2))/2*3

=> x = (-7 +/- (49 - 24)^(1/2))/6

=> x = (-7 +/- (25)^(1/2))/6

Now, we can get two solutions for root of 25, +5 and -5.

So,

x = (-7 + 5)/6 AND x = (-7 - 5)/6

=> x = -2/6 AND => x = -12/6

=> x = -1/3 AND => x = - 2(3 votes)

- Does the SAT require us to proof quadratics or not, if so, when do we use the proof?(2 votes)
- No, you will not be asked to show the proof for the quadratic formula on the SAT, although it is possible that you will have a question on the formula itself. This is not something that you use necessarily, but is rather a way of showing that it is valid.(5 votes)

## Video transcript

In the last video, I told you
that if you had a quadratic equation of the form ax squared
plus bx, plus c is equal to zero, you could use the
quadratic formula to find the solutions to
this equation. And the quadratic
formula was x. The solutions would be equal
to negative b plus or minus the square root of b squared
minus 4ac, all of that over 2a. And we learned how to use it. You literally just substitute
the numbers a for a, b for b, c for c, and then it gives you
two answers, because you have a plus or a minus right there. What I want to do
in this video is actually prove it to you. Prove that using, essentially
completing the square, I can get from that to that
right over there. So the first thing I want to
do, so that I can start completing the square from this
point right here, is-- let me rewrite the equation
right here-- so we have ax-- let me do it in a different
color-- I have ax squared plus bx, plus c is equal to 0. So the first I want to do is
divide everything by a, so I just have a 1 out here
as a coefficient. So you divide everything by a,
you get x squared plus b over ax, plus c over a, is equal
to 0 over a, which is still just 0. Now we want to-- well, let me
get the c over a term on to the right-hand side,
so let's subtract c over a from both sides. And we get x squared plus b over
a x, plus-- well, I'll just leave it blank there,
because this is gone now; we subtracted it from both sides--
is equal to negative c over a I left a space
there so that we can complete the square. And you saw in the completing
the square video, you literally just take 1/2 of this
coefficient right here and you square it. So what is b over
a divided by 2? Or what is 1/2 times b over a? Well, that is just b over 2a,
and, of course, we are going to square it. You take 1/2 of this
and you square it. That's what we do in completing
a square, so that we can turn this into the
perfect square of a binomial. Now, of course, we cannot just
add the b over 2a squared to the left-hand side. We have to add it
to both sides. So you have a plus b over 2a
squared there as well. Now what happens? Well, this over here, this
expression right over here, this is the exact same thing as
x plus b over 2a squared. And if you don't believe me, I'm
going to multiply it out. That x plus b over 2a squared
is x plus b over 2a, times x plus b over 2a. x times
x is x squared. x times b over 2a is
plus b over 2ax. You have b over 2a times x,
which is another b over 2ax, and then you have b over 2a
times b over 2a, that is plus b over 2a squared. That and this are the same
thing, because these two middle terms, b over 2a plus b
over 2a, that's the same thing as 2b over 2ax, which is the
same thing as b over ax. So this simplifies to x squared
plus b over ax, plus b over 2a squared, which is
exactly what we have written right there. That was the whole point of
adding this term to both sides, so it becomes
a perfect square. So the left-hand side
simplifies to this. The right-hand side, maybe
not quite as simple. Maybe we'll leave it the
way it is right now. Actually, let's simplify
it a little bit. So the right-hand side,
we can rewrite it. This is going to be equal
to-- well, this is going to be b squared. I'll write that term first. This
is b-- let me do it in green so we can follow along. So that term right there
can be written as b squared over 4a square. And what's this term? What would that become? This would become-- in order
to have 4a squared as the denominator, we have to multiply
the numerator and the denominator by 4a. So this term right
here will become minus 4ac over 4a squared. And you can verify for yourself
that that is the same thing as that. I just multiplied the
numerator and the denominator by 4a. In fact, the 4's cancel out and
then this a cancels out and you just have a c over a. So these, this and that
are equivalent. I just switched which I write
first. And you might already be seeing the beginnings of the
quadratic formula here. So this I can rewrite. This I can rewrite. The right-hand side, right
here, I can rewrite as b squared minus 4ac, all of
that over 4a squared. This is looking very close. Notice, b squared minus 4ac,
it's already appearing. We don't have a square root yet,
but we haven't taken the square root of both
sides of this equation, so let's do that. So if you take the square root
of both sides, the left-hand side will just become x plus--
let me scroll down a little bit-- x plus b over 2a is going
to be equal to the plus or minus square root
of this thing. And the square root of this
is the square root of the numerator over the square
root of the denominator. So it's going to be the plus or
minus the square root of b squared minus 4ac over the
square root of 4a squared. Now, what is the square
root of 4a squared? It is 2a, right? 2a squared is 4a squared. The square root of this
is that right here. So to go from here to here, I
just took the square root of both sides of this equation. Now, this is looking very
close to the quadratic. We have a b squared minus
4ac over 2a, now we just essentially have to subtract
this b over 2a from both sides of the equation and
we're done. So let's do that. So if you subtract the b over
2a from both sides of this equation, what do you get? You get x is equal to negative
b over 2a, plus or minus the square root of b squared minus
4ac over 2a, common denominator. So this is equal
to negative b. Let me do this in a new color. So it's orange. Negative b plus or minus the
square root of b squared minus 4ac, all of that over 2a. And we are done! By completing the square with
just general coefficients in front of our a, b and c, we
were able to derive the quadratic formula. Just like that. Hopefully you found that as
entertaining as I did.