Solving this for Theta confuses the heck out of me and its a rather simple equation $\displaystyle (1-2u\theta-\theta^2)$
Last edited by mr fantastic; Apr 9th 2009 at 03:29 AM. Reason: ll --> ck
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Originally Posted by sk8erboyla2004 Solving this for Theta confuses the hell out of me and its a rather simple equation $\displaystyle (1-2u\theta-\theta^2)$ 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
Yes it is equal to 0
Originally Posted by sk8erboyla2004 Yes it is equal to 0 You can use the quadratic formula to solve this. $\displaystyle -\theta^2 - 2u\theta +1 = 0$ compare to $\displaystyle ax^2+bx+c = 0$ I get an answer of $\displaystyle \theta = -u\pm \sqrt{u^2+1}$
Last edited by e^(i*pi); Apr 8th 2009 at 06:53 PM. Reason: oops forgot the minus sign
I get $\displaystyle \frac{2u\pm\sqrt{2u^2-4}}{-2}$ Im confused how to simplfy I can see how to reduce too $\displaystyle {-u\pm\sqrt{u^2-2}}$
Originally Posted by sk8erboyla2004 I get $\displaystyle \frac{2u\pm\sqrt{{\color{red}(-}2u{\color{red})}^2-4{\color{red}(-1)(1)}}}{-2}$ Im confused how to simplfy I can see how to reduce too $\displaystyle {-u\pm\sqrt{u^2-2}}$ You need to take more care. See my above edits in red and continue.
Originally Posted by sk8erboyla2004 I get $\displaystyle \frac{2u\pm\sqrt{2u^2-4}}{-2}$ Im confused how to simplfy I can see how to reduce too $\displaystyle {-u\pm\sqrt{u^2-2}}$ $\displaystyle {-u\pm\sqrt{u^2+2}}$
Last edited by mr fantastic; Apr 9th 2009 at 04:46 AM. Reason: Fixed latex
Originally Posted by sk8erboyla2004 $\displaystyle {-u\pm\sqrt{u^2+2}}$ No. Show your working. (The correct answer is given in post #4).
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