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find the real solutions for t:36t^4 + 33t^2 – 3 = 0

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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