vector a = (2,5) and vector b = (-1,t)
solve for t when the vectors have a 45 degree angle between them
any help would be appreciated.
$\displaystyle -2+5t = \dfrac{\sqrt{58(1+t^2)}}{2}$
$\displaystyle 10t-4 = \sqrt{58+58t^2}$
$\displaystyle 100t^2-80t+16 = 58+58t^2$
$\displaystyle 42t^2-80t-42 = 0$
$\displaystyle 21t^2-40t-21 = 0$
This one does factor but it might be easier to use the quadratic formula
Yes, that is the positive answer you get when you solve the quadratic.
$\displaystyle 21t^2 - 40 t - 21 = 0$
$\displaystyle t = \displaystyle\frac{40 \pm \sqrt{(-40)^ 2- 4(21)(-21)}}{2\times 21}$
$\displaystyle = \displaystyle\frac{40 \pm 58}{42}$