If r and s are the roots of ax^2+bx+c=0. (a, b, c are not equal to 0)?

Question

Then r^-1+s^-1 equals?

Answer 1

If r and s are the roots of ax² + bx + c = 0, then we can rewrite it as
x = r or x = s.

Transposing, we get
x - r = 0 or x - s = 0

Then
(x - r)(x - s) = 0

Multiply
x² - (r + s)x + rs = 0

We divide a from the quadratic
x² + b/a x + c/a = 0

Therefore,
r + s = -b/a and rs = c/a

Now, we need to find
r^-1 + s^-1

Since an exponent of -1 means the reciprocal, then
= 1/r + 1/s

Getting the LCD,
= (r + s)/rs

Now we substitute:
= (-b/a)/(c/a)

We cancel the a's:
= -b/c

Therefore,
r^-1 + s^-1 = -b/c

^_^

Answer 2

You want to find 1/r + 1/s. But this equals (r + s)/rs.
Now the sum of the roots of this equation is -b/a
and the product of the roots is c/a,
So the answer is -b/a / c/a = -b/c.

Answer 3

We have a(x-r)(x-s) = ax^2 + bx + c.
Expanding that gives ax^2 - a(r+s)x + a(rs) = 0.
So we know that -a(r+s) = b, so r + s = -b/a.
We also know that a(rs) = c, so rs = c/a.
So 1/r + 1/s = (r+s)/rs = (-b/a)/(c/a) = -b/c.

Answer 4

From http://en.wikipedia.org/wiki/Quadratic_equation#Vi.C3.A8te.27s_formulas we know that r+s=-b/a and r*s=c/a.
So we have 1/r+1/s=(r+s)/(r*s) = -b/c