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Sagot :
You have the following equation:
[tex]36t^4+33t^2-3=0[/tex]In order to solve for t, first divide by 3 both sides:
[tex]12t^4+11t^2-1=0[/tex]use the quadratic equation for t^2, as follow:
[tex]\begin{gathered} t^2=\frac{-11\pm\sqrt[]{(11)^2-4(12)(-1)}}{2(12)} \\ t^2=\frac{-11\pm13}{24} \\ t^2=\frac{-11+13}{24}=\frac{2}{24}=\frac{1}{12} \\ t^2=\frac{-11-13}{24}=\frac{-24}{24}=-1 \end{gathered}[/tex]If you only consider real solutions for t, then, you obtain two real solutions from t^2=1/12.
Then, the solutions are:
[tex]t=\frac{1}{\sqrt[]{12}},t=-\frac{1}{\sqrt[]{12}}[/tex]
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