Algebra (all content)
- 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
A text-based proof (not video) of the quadratic formula
The quadratic formula says that
for any quadratic equation like:
If you've never seen this formula proven before, you might like to watch a video proof, but if you're just reviewing or prefer a text-based proof, here it is:
We'll start with the general form of the equation and do a whole bunch of algebra to solve for . At the heart of the proof is the technique called . If you're unfamiliar with this technique, you may want to brush up by watching a video.
Part 1: Completing the square
Part 2: Algebra! Algebra! Algebra!
Remember, our goal is to solve for .
And we're done!
Want to join the conversation?
- When we take the sqrt of 4a^2 shouldn't it be + or - 2a, not just 2a?(17 votes)
- When you took the square root of the entire fraction, technically both sides have the 'plus or minus' symbol but putting it on both the numerator and the denominator would be redundant and you can simply simplify it like they did with a single plus or minus symbol in front of the fraction but you can also leave it on the numerator.(2 votes)
- In step 8 the square root on the right hand side is +/-. Why is the square root on the left hand side not also +/-?(9 votes)
- How can we multiply by 4a^2 in step 6, without affecting the left side of the equation?(4 votes)
- What they did in step 6 was multiply - c/a by 4a/4a. The reason you can do this is because 4a/4a is the same thing as 1, so multiplying by it doesn't change any values.
In other words, -c/a has the same value as -4ac/(4a^2)(5 votes)
- I require some help with understanding how -b/2a derives the x-coordinate of the vertex of a parabola. Thanks in advance!(1 vote)
- I know of two ways to understanding it.
First, using the vertex formula: y = a(x – h)^2 + k, where "h" is the vertex.
Put the general equation y = ax^2 + bx + c into the vertex form and you will find that "h" will equal -b/2a. I'll leave the work up to you.
Second, since quadratics in the general form (y = ax^2 + bx + c) are symmetric over a vertical line through the vertex, we can use the two roots of the quadratic formula and average them to find the x-coordinate of the vertex (visualize a quadratic graph and you will see why this is true).
So if you find the average of the two roots:
[-b + sqrt(b^2-4ac)]/2a and [-b - sqrt(b^2-4ac)]/2a // it will be -b/2a. (I again, will leave the work up to you.)(11 votes)
- can someone help me with 4x^2+11x-20=0 I solved everything expect I got stuck on the square root of 441(2 votes)
- You need to break 441 down into prime factors to simplify the square root.
Learn the divisibility tests -- see the video at this link: https://www.khanacademy.org/math/pre-algebra/pre-algebra-factors-multiples/pre-algebra-divisibility-tests/v/divisibility-tests-for-2-3-4-5-6-9-10
If you know these test, you would recognize that 441 is divisible by 9.
441 / 9 = 49
So, prime factors of 441 = 3 * 3 * 7 * 7
And sqrt(441) = 21
Hope this helps.(4 votes)
- I tried the proof myself in a slightly different way and it didn't quite work out.
(1) ax^2 + bx + c = 0
(2) x^2 + (b/a)x + c/a = 0
(3) x^2 + (b/a)x + (b^2/4a^2) + c/a - (b^2/4a^2) = 0
(4) (x+(b/2a))^2 + 4ac/4a^2 - b^2/4a^2 = 0
(5) (x+(b/2a))^2 + (4ac-b^2)/(4a^2) = 0
(6) (x+(b/2a))^2 = -(4ac-b^2)/(4a^2)
(7) (x+(b/2a)) = -(sqrt(4ac-b^2))/2a
(8) x = -(b/2a)-(sqrt(4ac-b^2))/2a
(9) x = -(-b+-squr(4ac-b^2))/2a
My discriminant is 4ac-b^2 while the one we use is b^2-4ac. Also, I have a negative sign in front of the whole fraction which is from the second equation in step 8. Could somebody tell me where I made a mistake?(3 votes)
- At step 6 and 7 when you took the square root, the negative should have stayed inside rather than outside (taking square root would yield ± on the outside). So when you distribute the -1(4ac-b^2) you end up with b^2-4ac.(1 vote)
- on step 4, why adding (b^2/4a^2) to both sides?(1 vote)
- To complete the square, you divide the coefficient of the x term by 2 (b/2a) and square this to get b^2/4a^2. So you need this term to complete the square. If you do it to the left side in order to complete the square, you either have to subtract it on the left or add it to the right side of the equation to keep it balanced.(5 votes)
- Why did you add +/-(1 vote)
- Do you mean in step 8? This is because 2^2 is 4 AND (-2)^2 is ALSO 4. Therefore, the square root could be both positive and negative, so we add a +/- sign.(4 votes)
- n step 8 the square root on the right hand side is +/-. Why is the square root on the left hand side not also +/-?(2 votes)
- That is a great question! I thought I knew algebra, but I never noticed that and it took me a little minute to work out!
Only one side of the equation needs a +/- sign because if you multiply the equation by -1 you can get to any combination of negatives and positives while only putting the +/- sign on one side. I know that sounds confusing, so let's simplify things. Say, for example, we're using the equation a = ±b. The two possible situations are a = b and a = -b, but multiplying those by -1 gives you -a = -b and -a = b, meaning that either side could be positive or negative in any combination with the +/- sign on only one side. Terrific question! I hope this helps, and remember that you can learn anything!(2 votes)