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Use the quadratic formula for (t^2 - 3)^2.
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I got the answer in decimals by factoring by my teacher said it has to be in radical form.
are you trying to find the roots of \[(t^2-3)^2\] i.e. finding the solutions to \[(t^2-3)^2=0\] ?
Yes, the zeros (roots).
ok \[(t^2-3)^2=0\] take square root \[(t^2-3)=\sqrt{0}=0\] add 3 \[t^2=3\] take square root again \[t=\pm\sqrt{3}\]
And what would be the two zeros (or roots)? Can you go any farther and not have decimals?
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that is it \[t=\sqrt{3}\] and \[t=-\sqrt{3}\]
Oh! Wow, I can't believe I never understood that. Thank you, you just explained a bunch to me.
no problem
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