Prove that if the diffrence between the root of the equation ax^2+bx+c=0 is 1, then a=b^2-a^2/4c
the roots would be
(-b +√(b^2 - 4ac)/2a and (-b - √(b^2 -4ac)/2a
we are told that the difference between them is 1
so,
(-b+√(b^2-4a))/2a - (-b - √(b^2-4ac)/2a = 1
2√(b^2-4ac)/2a = 1
√(b^2 - 4ac) = a
square both sides
b^2 - 4ac = a^2
b^2 - a^2 = 4ac
divide both sides by 4c and we have our result
(b^2 - a^2)/4c = a