1. ## Multiplying by conjugate

4-sqrt(s)/(s-16)

s=16

i keep multiplying this by 4+sqrt(s) and keep coming up with -12/0
the book is telling me that it's -1/8... how is this...

2. Originally Posted by NotEinstein
4-sqrt(s)/(s-16)

s=16

i keep multiplying this by 4+sqrt(s) and keep coming up with -12/0
the book is telling me that it's -1/8... how is this...
Did you factorise the numerator ?

3. Originally Posted by Moo
Did you factorise the numerator ?
I'm all jumbled right now.. now i'm getting 0/0... *cries*

5. $\displaystyle 4-sqrt(s)/(s-16) * (4+sqrt(s)/4+sqrt(s))$

$\displaystyle 16-s/(s-16)(4+sqrt(s))$

$\displaystyle 0/ (4s+s(sqrt(s))-64-(16(sqrt(s)))$

$\displaystyle 0/ (64 + 64 - 64 - 64)$

0/0...

6. I fixed your Latex code. Take a look so you know how to write fractions now

Oh no! You changed it back to wrong.... sigh. When writing fractions in Latex, use \frac{numerator}{denominator}

I spotted your error. You don't need to plug in s=16 and get 0 in the numerator. 16-s and s-16 are only a negative sign factored out away from equal.

$\displaystyle \frac{16-s}{(s-16)(4+sqrt(s))} = \frac{16-s}{-(16-s)(4+\sqrt{s})}$.

Now the 16-s cancels and you are left with $\displaystyle -\frac{1}{4+\sqrt{s}}$

7. $\displaystyle \frac{4-\sqrt{s}}{s-16} \bigg(\frac{4+\sqrt{s}}{4+\sqrt{s}}\bigg)$

$\displaystyle \frac{16-s}{s-16}\bigg(\frac{1}{4+\sqrt{s}}\bigg)$

$\displaystyle \frac{0}{4s+s\sqrt{s}-64-16\sqrt{s}}$

8. Originally Posted by Jameson
I fixed your Latex code. Take a look so you know how to write fractions now

Oh no! You changed it back to wrong.... sigh. When writing fractions in Latex, use \frac{numerator}{denominator}

I spotted your error. You don't need to plug in s=16 and get 0 in the numerator. 16-s and s-16 are only a negative sign factored out away from equal.

$\displaystyle \frac{16-s}{(s-16)(4+sqrt(s))} = \frac{16-s}{-(16-s)(4+\sqrt{s})}$.

Now the 16-s cancels and you are left with $\displaystyle -\frac{1}{4+\sqrt{s}}$

so you can just do that? just toss in a negative like that? geeezz... no algebra for 5 years really messes with my head

9. I wouldn't call it "tossing in a negative".

Here, think about it. Let's say you have (a-b). Now if you factor out a -1 from both terms, what do you get? -1(-a+b) or -1(b-a) or -(b-a). If you distribute the -1 back through both terms you get your original sum.