Question

If the roots of ax^2 + bx + c=0 differ by 1, show that they are `(a-b)/(2a)` and `- (a+b)/(2a)` . and prove that `b^2 = a(a+4a)`?

Answer 1

Please see below.

Explanation:

Using either completing the square or the quadratic formula, we get solutions

`x = (-b+-sqrt(b^2-4ac))/(2a)`

We are told that the difference in the solutions is `1`, so we have

`(-b+sqrt(b^2-4ac))/(2a)-((-b-sqrt(b^2-4ac))/(2a))=1`.

So, `(2sqrt(b^2-4ac))/(2a) = 1`.

Hence, `b^2-4ac=a^2`

This gets us the final equality `b^2=a^2-4ac`

and it also allows us to replace `b^2-4ac` with `a^2` to get

`x = (-b+-sqrt(a^2))/(2a)`

So the solutions are

`x = (-b+-a)/(2a)` as required.