The essence of the proof is simply to complete the square for a generalised quadratic equation. Like this:
ax2 + bx + c = 0
Take 'a' outside:
a[x2 + bx/a + c/a] = 0
Divide through by 'a':
x2 + bx/a + c/a = 0
Complete the square:
(x + b/2a)2 - b2/4a2 + c/a = 0
Rearrange to find x:
(x + b/2a)2 = b2/4a2 - c/a
x + b/2a = (+/-)sqrt[b2/4a2 - c/a]
x = -b/2a (+/-) sqrt[b2/4a2 - c/a]
Finally, fiddle around so that (1/2a) can be taken out as a common factor:
x = -b/2a (+/-) sqrt[b2/4a2 - 4ac/4a2]
x = -b/2a (+/-) sqrt[(1/4a2)(b2 - 4ac)]
x = -b/2a (+/-) sqrt(1/4a2)sqrt(b2 - 4ac)
x = -b/2a (+/-) (1/2a)sqrt(b2 - 4ac)
x = [ -b (+/-) sqrt(b2 - 4ac) ] / 2a.
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There are an infinite number of different quadratic equations. The quadratic formula is a single formula that can be used to find the pair of solutions to every quadratic equation.
Well, if the given quadratic equation cannot be factored, nor completed by the square, try using the quadratic formula.
Quadratic formula. It's easier to remember and you have to do less work.
For any quadratic ax2 + bx + c = 0 we can find x by using the quadratic formulae: the quadratic formula is... [-b +- sqrt(b2 - 4(a)(c)) ] / 2a
The quadratic formula is x=-b (+or-) square root of b2-4ac all over 2a