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Quadratic formula proof review

A text-based proof (not video) of the quadratic formula
The quadratic formula says that
x, equals, start fraction, minus, start color #e07d10, b, end color #e07d10, plus minus, square root of, start color #e07d10, b, end color #e07d10, squared, minus, 4, start color #7854ab, a, end color #7854ab, start color #e84d39, c, end color #e84d39, end square root, divided by, 2, start color #7854ab, a, end color #7854ab, end fraction
for any quadratic equation like:
start color #7854ab, a, end color #7854ab, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #e84d39, c, end color #e84d39, equals, 0
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:

The proof

We'll start with the general form of the equation and do a whole bunch of algebra to solve for x. At the heart of the proof is the technique called start color #11accd, start text, c, o, m, p, l, e, t, i, n, g, space, t, h, e, space, s, q, u, a, r, e, end text, end color #11accd. If you're unfamiliar with this technique, you may want to brush up by watching a video.

Part 1: Completing the square

ax2+bx+c=0(1)ax2+bx=c(2)x2+bax=ca(3)x2+bax+b24a2=b24a2ca(4)(x+b2a)2=b24a2ca(5)\begin{aligned} \purpleD{a}x^2 + \goldD{b}x + \redD{c} &= 0&(1)\\\\ ax^2+bx&=-c&(2)\\\\ x^2+\dfrac{b}{a}x&=-\dfrac{c}{a}&(3)\\\\ \blueD{x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(4)\\\\ \blueD{\left (x+\dfrac{b}{2a}\right )^2}&\blueD{=\dfrac{b^2}{4a^2}-\dfrac{c}{a}}&(5) \end{aligned}

Part 2: Algebra! Algebra! Algebra!

Remember, our goal is to solve for x.
(x+b2a)2=b24a2ca(5)(x+b2a)2=b24a24ac4a2(6)(x+b2a)2=b24ac4a2(7)x+b2a=±b24ac4a2(8)x+b2a=±b24ac2a(9)x=b2a±b24ac2a(10)x=b±b24ac2a(11)\begin{aligned} \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{c}{a}&(5) \\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2}{4a^2}-\dfrac{4ac}{4a^2} &(6)\\\\ \left (x+\dfrac{b}{2a}\right )^2&=\dfrac{b^2-4ac}{4a^2}&(7)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{\sqrt{4a^2}}&(8)\\\\ x+\dfrac{b}{2a}&=\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(9)\\\\ x&=-\dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a}&(10)\\\\ x&=\dfrac{-\goldD{b}\pm\sqrt{\goldD{b}^2-4\purpleD{a}\redD{c}}}{2\purpleD{a}}&(11) \end{aligned}
And we're done!

Want to join the conversation?

  • duskpin ultimate style avatar for user jd1311993
    When we take the sqrt of 4a^2 shouldn't it be + or - 2a, not just 2a?
    (17 votes)
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    • mr pink red style avatar for user hemaltavleen
      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)
  • spunky sam orange style avatar for user Derrick Logan
    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)
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  • aqualine ultimate style avatar for user PythagorasLessFortunateBrother
    How can we multiply by 4a^2 in step 6, without affecting the left side of the equation?
    (4 votes)
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  • leaf orange style avatar for user Raghav
    I require some help with understanding how -b/2a derives the x-coordinate of the vertex of a parabola. Thanks in advance!
    (1 vote)
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    • piceratops seedling style avatar for user bsd
      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)
  • aqualine seed style avatar for user ranoosh
    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)
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  • leaf green style avatar for user Lyn Kang
    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)
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  • leaf red style avatar for user Hann
    on step 4, why adding (b^2/4a^2) to both sides?
    (1 vote)
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    • mr pink green style avatar for user David Severin
      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)
  • piceratops seedling style avatar for user Hrithik
    Why did you add +/-
    (1 vote)
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  • blobby green style avatar for user Santiago Ramirez
    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)
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    • leafers ultimate style avatar for user Neil Gabrielson
      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)
  • duskpin ultimate style avatar for user Jerusha Curlin
    How is -4ac/4a^2 equal to -c/a?
    (2 votes)
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