the proof is simply to complete the square for a generalised quadratic equation. Like this:
ax2(xsquared) + 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.
how do you prove the quadratic equation?
2 answers
a bit easier to read ....
divide by a
x^2 + (b/a)x = -c/a
complete the square , add b^2/(2a^2) to both sides
x^2 + (b/a)x + b^2/(4a^2) = b^2/(4a^2) - c/a
write the left side as a square and add the two terms on the right
(x + b/(2a) )^2 = (b^2 - 4ac)/(4a^2)
take √ of both sides
x + b/(2a) = ± √(b^2 - 4ac)/(2a)
x = -b/(2a) ± √(b^2 - 4ac)/(2a)
= (-b ± √(b^2 - 4ac)/(2a)
divide by a
x^2 + (b/a)x = -c/a
complete the square , add b^2/(2a^2) to both sides
x^2 + (b/a)x + b^2/(4a^2) = b^2/(4a^2) - c/a
write the left side as a square and add the two terms on the right
(x + b/(2a) )^2 = (b^2 - 4ac)/(4a^2)
take √ of both sides
x + b/(2a) = ± √(b^2 - 4ac)/(2a)
x = -b/(2a) ± √(b^2 - 4ac)/(2a)
= (-b ± √(b^2 - 4ac)/(2a)