How can you rearrange the equation to get a positive tension at the front?

[RK1992]

Given:

$$T=mgcos\theta$$

and

$$cos^2\theta + \frac{mV^2}{ag} cos\theta - 1 = 0$$

show that:

$$T = \frac{mV^2}{2a} [1+\sqrt{1+(\frac{2ag}{V^2})^2}]$$

I always get:

$$T = - \frac{mV^2}{2a} [1 + \sqrt{1 + (\frac{2ag}{V^2})^2}]$$

T represents a tension in a string so I understand why you'd take just the positive root (can't have a negative tension in a string) but I don't see how you can get a positive mV²/2a out in the front.. Any ideas if it's me who's wrong or the question? If it is impossible to rearrange so there's not a negative outside, do you just take the magnitude for the reason I stated, because it's a string?

 

[jbsteele]

The answer given is correct. To get it, you should solve the equation for cosθ:cos^2\theta + \frac{mV^2}{ag} cos\theta - 1 = 0 cos\theta = \frac{-\frac{mV^2}{ag} \pm \sqrt{\frac{m^2V^4}{a^2g^2}+ 4}}{2} Substitute this expression into the formula for T to get:T = mg\frac{-\frac{mV^2}{ag} \pm \sqrt{\frac{m^2V^4}{a^2g^2}+ 4}}{2} Simplify to get:T = \frac{mV^2}{2a} \pm \frac{mV^2}{2a}\sqrt{1 + \frac{4a^2g^2}{m^2V^4}} Since we cannot have a negative tension in a string, the negative root must be discarded and you are left with:T = \frac{mV^2}{2a} [1+\sqrt{1+(\frac{2ag}{V^2})^2}]