# Thread: Solving This Variable confuse me

1. ## Solving This Variable confuse me

Solving this for Theta confuses the heck out of me and its a rather simple equation

$(1-2u\theta-\theta^2)$

2. Originally Posted by sk8erboyla2004
Solving this for Theta confuses the hell out of me and its a rather simple equation

$(1-2u\theta-\theta^2)$
You need it to be equal to something to solve as an equation. If it is equal to 0 use the quadratic formula with a= -1, b = -2u and c =1

3. Yes it is equal to 0

4. Originally Posted by sk8erboyla2004
Yes it is equal to 0
You can use the quadratic formula to solve this.

$-\theta^2 - 2u\theta +1 = 0$

compare to $ax^2+bx+c = 0$

I get an answer of $\theta = -u\pm \sqrt{u^2+1}$

5. I get

$\frac{2u\pm\sqrt{2u^2-4}}{-2}$

Im confused how to simplfy

I can see how to reduce too

${-u\pm\sqrt{u^2-2}}$

6. Originally Posted by sk8erboyla2004
I get

$\frac{2u\pm\sqrt{{\color{red}(-}2u{\color{red})}^2-4{\color{red}(-1)(1)}}}{-2}$

Im confused how to simplfy

I can see how to reduce too

${-u\pm\sqrt{u^2-2}}$
You need to take more care. See my above edits in red and continue.

7. Originally Posted by sk8erboyla2004
I get

$\frac{2u\pm\sqrt{2u^2-4}}{-2}$

Im confused how to simplfy

I can see how to reduce too

${-u\pm\sqrt{u^2-2}}$

${-u\pm\sqrt{u^2+2}}$

8. Originally Posted by sk8erboyla2004
${-u\pm\sqrt{u^2+2}}$
No. Show your working. (The correct answer is given in post #4).