1. ## Solve for W

Solve for W

$\displaystyle V=\sqrt{2Lw+w^2}$
My answer was $\displaystyle \Longrightarrow w=V^2-2L$
But apparently thats wrong according to my teacher.

2. Originally Posted by OzzMan
Solve for W

$\displaystyle V=\sqrt{2Lw+w^2}$
My answer was $\displaystyle \Longrightarrow w=V^2-2L$
But apparently thats wrong according to my teacher.
Square both sides of the equation:

$\displaystyle V^2=2Lw + w^2~\iff~w^2 + 2Lw - V^2 = 0$

This is a quadratic equation. Use the formula to solve it.

3. What do you mean exactly? What I did was square both sides like you said and got $\displaystyle V^2=2Lw+w^2 \Longrightarrow w^2=V^2 - 2Lw$
Then I did $\displaystyle \frac{w^2}{w}=V^2 - \frac{2Lw}{w}$

4. Originally Posted by OzzMan
What do you mean exactly? What I did was square both sides like you said and got $\displaystyle V^2=2Lw+w^2 \Longrightarrow w^2=V^2 - 2Lw$
Then I did $\displaystyle \frac{w^2}{w}=V^2 - \frac{2Lw}{w}$
Unfortunately you forgot to divide the Vē by w.

If you have a quadratic equation :

$\displaystyle ax^2+bx+c = 0$

Then this equation has the solutions:

$\displaystyle w = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

a = 1
b = -2L
c = -Vē

Plug in these terms into the formula to get the solutions.

(For confirmation only: $\displaystyle w = L\pm\sqrt{L^2+V^2}$

5. Just making sure I did this right
$\displaystyle w=\frac{2L\pm\sqrt{4L^2+4v^2}}{2}$
How did you get $\displaystyle w=L \pm\sqrt{L^2+V^2}$

6. $\displaystyle \sqrt {4L^2 + 4w^2 } = \sqrt {4\left( {L^2 + w^2 } \right)} = \sqrt 4 \sqrt {\left( {L^2 + w^2 } \right)}$

7. shouldn't it be a $\displaystyle -2L$ instead of $\displaystyle 2L$? According to the equation $\displaystyle w^2+2Lw-V^2=0$
$\displaystyle w=\frac{-2L\pm\sqrt{4L^2+4v^2}}{2}$
So the answer would be $\displaystyle w = -L\pm\sqrt{L^2+V^2}$

8. Originally Posted by OzzMan
shouldn't it be a -2L instead of 2L? According to the equation $\displaystyle w^2+2Lw-V^2=0$
$\displaystyle w=\frac{-2L\pm\sqrt{4L^2+4v^2}}{2}$
So the answer would be $\displaystyle w = -L\pm\sqrt{L^2+V^2}$
Yes. Good pick-up. Earboth made a small sign error when he substituted for b:

Originally Posted by earboth
[snip]
(For confirmation only: $\displaystyle w =$ L$\displaystyle \pm\sqrt{L^2+V^2}$