Question

solve:t^4 +12t^2 =32

Answer 1

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$= t {}^{4} + 12t {}^{2} = 32$

$= t {}^{4} + 12t {}^{2} - 32 = 0$

$= u {}^{2} + 12u - 32 = 0$

$=t = - 6 + 2 \sqrt{17}$

$= t = - 6 - 2 \sqrt{17}$

$= t1 = - \sqrt{ - 6 + 2 \sqrt{17} } \\ = t2 = \sqrt{ - 6 + 2 \sqrt{17} }$

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