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# Proof of the quadratic formula

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? •  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.
• At , 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? • | 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. :(
• At , why did he simply not just take the square root of b squared and leave it as b? • At , why can't you bring the b^2 out from under the radical?

The square root of b^2 is b, right? • 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).
• At around , 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? • Is there a similar formula for the solutions of a cubic equation? • Anyone with the question "Where is the Quadratic Formula video that he references in this video?"
It's the first video in the next unit. I can sort of see why they did it in this order because here they are just showing you where it comes from, you don't need to know anything about it yet, so the order it is in currently is probably better than the original order. • 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 • 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!
• What does vertex form of a quadratic equation means? And is this right:
a(x-h)²+k=0
x=h±√-k/a • how do you find 2 solutions • 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

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