# Math Help - solve for s

1. ## solve for s

1/s+a + 1/s+b = 1/c

solve for s

2. Originally Posted by aeroflix
1/s+a + 1/s+b = 1/c

solve for s
Is it $\dfrac{1}{s}+a+\dfrac{1}{s}+b=\dfrac{1}{c}$ or $\dfrac{1}{s+a}+\dfrac{1}{s+b}=\dfrac{1}{c}$? Depending on the interpretation, you will get completely different answers!!

3. Originally Posted by aeroflix
1/s+a + 1/s+b = 1/c
solve for s
IF: 1/(s+a) + 1/(s+b) = 1/c
THEN:
s = {2c - a - b +- SQRT[(a + b - 2c)^2 - 4(ab - ac - bc)]} / 2

4. Originally Posted by aeroflix
1/s+a + 1/s+b = 1/c

solve for s
$\frac{1}{s + a} + \frac{1}{s + b} = \frac{1}{c}$

$\frac{s + b}{(s + a)(s + b)} + \frac{s + a}{(s + a)(s + b)} = \frac{1}{c}$

$\frac{s + b + s + a}{(s + a)(s + b)} = \frac{1}{c}$

$\frac{2s + a + b}{s^2 + (a + b)s + ab} = \frac{1}{c}$

$c(2s + a + b) = s^2 + (a + b)s + ab$

$2cs + ac + bc = s^2 + (a + b)s + ab$

$0 = s^2 + (a + b - 2c)s + ab - ac - bc$.

Now solve using the Quadratic Formula...

5. whats the value of c ???

value of a is 1 and b is what also?

6. You need to solve for s in terms of a, b, and c. The values of a, b, and c can be taken to be arbitrary constants (ie, they don't change, but you also don't know what they are).

7. ive came up with a quadratic but the part -4ac kinda freaks me out

8. Originally Posted by aeroflix
whats the value of c ???

value of a is 1 and b is what also?
Originally Posted by Math Major
You need to solve for s in terms of a, b, and c. The values of a, b, and c can be taken to be arbitrary constants (ie, they don't change, but you also don't know what they are).
No, the OP is asking what is a, b, and c in the quadratic formula. So in the equation
$s^2 + (a + b - 2c)s + ab - ac - bc = 0$
"a" is 1,
"b" is a + b - 2c, and
"c" is ab - ac - bc.

So the setup would be
$s = \frac{-(a + b - 2c) \pm \sqrt{(a + b - 2c)^2 - 4(1)(ab - ac - bc)}}{2(1)}$
...