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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 is the word quadratic used describe a 2nd degree polynomail, I suspect has to do with geomerty and the square. Perhaps you can make a video about how squares can be used to represent a second degree polynomail. Nice proof.
• The adjective quadratic comes from the Latin word quadratum for square. A term like x^2 is called a square in algebra because it is the area of a square with side x.

In general, a prefix quadr(i)- indicates the number 4. Examples are quadrilateral and quadrant. Quadratum is the Latin word for square because a square has four sides.

• How do I use the quadratic equation to find the formula for the vertex? Thanks Sal!
• when you look on the quadratic formula x= [-b+/-sqrt(b^2-4ac)]/2a for the vertex all you need is this x= [-b/2a] and just plug it in.
• 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?
• 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.
• 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?
• 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.
• Is there a similar formula for the solutions of a cubic equation?
• 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:

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.
• 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!