1. ## algebra

if $\frac{a}{b}=\frac{1+p}{1-p}$, find the value of $\frac{a^2+b^2}{a^2-b^2}$ in terms of p.

2. Hello,
Originally Posted by requal
if $\frac{a}{b}=\frac{1+p}{1-p}$, find the value of $\frac{a^2+b^2}{a^2-b^2}$ in terms of p.
$\frac{a^2+b^2}{a^2-b^2}=\frac{b^2 \left(\frac{a^2}{b^2}+1\right)}{b^2 \left(\frac{a^2}{b^2}-1\right)}=\frac{\left(\frac ab\right)^2+1}{\left(\frac ab\right)^2-1}$

Better now ?

Otherwise, you could write $a=b \cdot \frac{1+p}{1-p}$ and just substitute. There would have been a common factor : $b^2$

3. Hello, requal!

Another approach . . .

$\text{If }\,\frac{a}{b}=\frac{1+p}{1-p},\:\text{ find the value of: }\:\frac{a^2+b^2}{a^2-b^2}\,\text{ in terms of }p.$
We have: . $\frac{a}{b} \:=\:\frac{1+p}{1-p} \quad\Rightarrow\quad \frac{a^2}{b^2} \:=\:\frac{1+2p + p^2}{1-2p+p^2}$ .[1]

Add 1 to both sides of [1]: . $\frac{a^2}{b^2}+1 \:=\:\frac{1+2p+p^2}{1-2p+p^2} + 1$

. . and we have: . $\frac{a^2+b^2}{b^2} \:=\:\frac{2+2p^2}{1-2p+p^2} \quad\Rightarrow\quad a^2+b^2\:=\:\frac{2b^2(1+p^2)}{(1-p)^2}$ .[2]

Subtract 1 from both sides of [1]: . $\frac{a^2}{b^2}-1 \:=\:\frac{1+2p+p^2}{1-2p+p^2} - 1$

. . and we have: . $\frac{a^2-b^2}{b^2} \:=\:\frac{4p}{1-2p+p^2} \quad\Rightarrow\quad a^2-b^2 \:=\:\frac{4b^2p}{(1-p)^2}$ .[3]

Divide [2] by [3]: . $\frac{a^2+b^2}{a^2-b^2} \;=\;\frac{\dfrac{2b^2(1+p^2)}{(1-p)^2}} {\dfrac{4b^2p}{(1-p)^2}} \;=\;\frac{1+p^2}{2p}\quad\hdots\:\text{ta-}DAA!$

4. It's pretty equivalent

Maybe yours can still be simplified.

Originally Posted by Soroban
. . and we have: . $\frac{a^2+b^2}{b^2} \:=\:\frac{2+2p^2}{1-2p+p^2} \quad\Rightarrow\quad a^2+b^2\:=\:\frac{2b^2(1+p^2)}{(1-p)^2}$ .[2]
Just keep $\frac{a^2+b^2}{b^2} \:=\:\frac{2+2p^2}{1-2p+p^2}$

. . and we have: . $\frac{a^2-b^2}{b^2} \:=\:\frac{4p}{1-2p+p^2} \quad\Rightarrow\quad a^2-b^2 \:=\:\frac{4b^2p}{(1-p)^2}$ .[3]
Keep $\frac{a^2-b^2}{b^2} \:=\:\frac{4p}{1-2p+p^2}$

Dividing the two equations, we directly find $\frac{a^2+b^2}{a^2-b^2}$

The problem is that maybe one can't see why we have to subtract/add 1 to a²/b²

5. Originally Posted by Moo
Hello,

$\frac{a^2+b^2}{a^2-b^2}=\frac{b^2 \left(\frac{a^2}{b^2}+1\right)}{b^2 \left(\frac{a^2}{b^2}-1\right)}=\frac{\left(\frac ab\right)^2+1}{\left(\frac ab\right)^2-1}$

Better now ?
Sorry to bump this thread but I don't follow? I was thinking you were trying to make it into 1+p/1-p but its thats in the form of 1+p/p+1?

6. Originally Posted by Soroban
Hello, requal!

Another approach . . .

We have: . $\frac{a}{b} \:=\:\frac{1+p}{1-p} \quad\Rightarrow\quad \frac{a^2}{b^2} \:=\:\frac{1+2p + p^2}{1-2p+p^2}$ .[1]

Add 1 to both sides of [1]: . $\frac{a^2}{b^2}+1 \:=\:\frac{1+2p+p^2}{1-2p+p^2} + 1$

. . and we have: . $\frac{a^2+b^2}{b^2} \:=\:\frac{2+2p^2}{1-2p+p^2} \quad\Rightarrow\quad a^2+b^2\:=\:\frac{2b^2(1+p^2)}{(1-p)^2}$ .[2]

Subtract 1 from both sides of [1]: . $\frac{a^2}{b^2}-1 \:=\:\frac{1+2p+p^2}{1-2p+p^2} - 1$

. . and we have: . $\frac{a^2-b^2}{b^2} \:=\:\frac{4p}{1-2p+p^2} \quad\Rightarrow\quad a^2-b^2 \:=\:\frac{4b^2p}{(1-p)^2}$ .[3]

Divide [2] by [3]: . $\frac{a^2+b^2}{a^2-b^2} \;=\;\frac{\dfrac{2b^2(1+p^2)}{(1-p)^2}} {\dfrac{4b^2p}{(1-p)^2}} \;=\;\frac{1+p^2}{2p}\quad\hdots\:\text{ta-}DAA!$

Is there some special reason that you would add/subtract one from both equations? Or is it just to make it easier to do

7. Originally Posted by Moo
Hello,

$\frac{a^2+b^2}{a^2-b^2}=\frac{b^2 \left(\frac{a^2}{b^2}+1\right)}{b^2 \left(\frac{a^2}{b^2}-1\right)}=\frac{\left(\frac ab\right)^2+1}{\left(\frac ab\right)^2-1}$

Better now ?
Originally Posted by requal
Sorry to bump this thread but I don't follow? I was thinking you were trying to make it into 1+p/1-p but its thats in the form of 1+p/p+1?
It is not bumping when you ask a follow-up question. Follow-up questions are always welcome and encouraged.

Substitute $\frac{a}{b} = \frac{1 + p}{1 - p}$ into $\frac{\left(\frac ab\right)^2+1}{\left(\frac ab\right)^2-1}$.