math proof help?

Question

Let a,b, and c be real #s with a not equal zero. Find error in the following proof that (-b/2a) is solution to ax^2+bx+c=0.

Let x and y be solutions to the equation. Subtracting ay^2 + by + c = 0 from ax^2 + bx + c = 0 yields a(x^2 + y^2) + b(x -y)=0, which we rewrite as a(x+y)(x-y)+b(x-y)=0. Hence a(x+y)+b=0, and thus x+y=-b/a. Since x and y can be any solutions, we can apply this computation letting y have the same value as x. With y=x, we obtain 2x=-b/a, or x=-b/2a.

I did this:
ax^2 + bx + c = 0
-[ay^2 + by + c = 0]
=ax^2 - ay^2 + bx - by + c - c = 0

a(x^2-y^2)+b(x-y)=0
a(x-y)(x+y)+b(x-y)=0
a(x+y)(x-y)=-b(x-y)
a(x+y)=-b
a(x+y)+b=0
x+y=-b/a

if y=x then
2x= -b/a or x= -b/2a

I dont see an error?

Answer 1

Last sentence: With y=x, we obtain 2x=-b/a, or x=-b/2a.

If y = x, then x - y = 0.

when they transitions from a(x+y)(x-y) +b(x - y) = 0, they divided both sides by (x - y). Dividing by 0 is BAD, well okay, it makes the equation invalid from this point onward, because division by zero is undefined.

Answer 2

The error is that you assumed x = y. If that were the case, the discriminant (b^2 - 4ac) would have been zero, so there would only be one root, -b+-0 over 2a (by quadratic formula).

But if the 2 roots are different it doesn't work.

And the above answers are right too. If you assume that if

a times 0 = b times 0, then a = b, you can prove any two numbers are equal.

Answer 3

1) The general solution to the quadratic is
x = (-b + k*sqrt( b^2 - 4ac )) / 2a; where k = plus or minus 1

2) Now, x1 = -b / (2a) is a solution only if b^2 - 4ac = 0;

3) So, the proof is wrong except for the special case where c = b^2 / (4a)

Answer 4

The error lies in this step:

a(x + y)(x - y) = -b(x - y)

You can't divide both sides by (x - y), because if x = y, then x - y = 0. You're not allowed to divide by 0.