Another cool thing I never mentioned that is cool about vertex form is that you can calculate x when the equation is in vertex form. For (x + 3)^2 – 1 = 0, you can say that x = -2 or -4, just by simply manipulating variables.
(x + 3)^2 – 1 = 0
(x + 3)^2 = 1 (add one to both sides)
x + 3 = 1 or -1 (take the ± square root of both sides; use 1 and -1 because 1
and -1 squared are both 1)
and -1 squared are both 1)
x = -2 or -4 (subtract three from both sides, from the 1 and the -1 to get
both answers)
both answers)
What about for standard form, using the coefficients a, b, and c. Maybe there is a formula we can derive just by following the steps involved in completing the square. Remember our original equation: ax2 + bx + c = 0. We’ll do it alongside an example utilizing every step, 2x2 – 4x + 6 = 0.
Starting Equation
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ax2 + bx + c = 0
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2x2 – 4x – 6 = 0
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Move the constant over to the right hand side, just to make things less messy.
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ax2 + bx = -c
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2x2 – 4x = 6
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Divide through by a. Normally, we would factor it, but it is the same thing.
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x2 + b/ax = -c/a
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x2 – 2x = 3
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Take half of the new b term and square it.
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(b/2a)2 = b^2/4a^2
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(-1)2 = 1
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Add in and subtract out this number. For this, we’ll just add it to both sides, since c is on the other side.
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x2 + b/ax + b^2/4a^2 = -c/a + b^2/4a^2
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x2 – 2x + 1 = 3 + 1
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Simplify the right hand side. For the first example, we will multiply the first fraction by 4a/4a to get a common denominator.
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4a/4a(-c/a) + b^2/4a^2
-4ac/4a^2 + b^2/4a^2
b^2 – 4ac/4a^2
x2 + b/ax + b^2/4a^2 = b^2 – 4ac/4a^2
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x2 – 2x + 1 = 4
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Convert the left hand side into squared form.
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(x + b/2a)2 = b^2 – 4ac/4a^2
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(x – 1)2 = 4
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Square root both sides. Don't forget to put our ± sign before the square root!
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x + b/2a = ±√(b^2 – 4ac)/2a
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x – 1 = ±2
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Isolate the x by adding/subtracting the remaining constant.
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x = -b/2a ± √(b^2 – 4ac)/2a
x = -b ± √(b^2 – 4ac)/2a
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x = 1 ± 2
x = -1 or 3 |
This formula is pretty well-known, but I think it's cool how it is derived. This is a formula that you want to memorize, because it is so important.
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