Answer
$\left\{-\sqrt{7},-\sqrt{2},\sqrt{2},\sqrt{7}\right\}$
Work Step by Step
In the form $ax^2+bx+c=0,$ the given equation, $
t^4+14=9t^2
,$ is equivalent to
\begin{align*}
t^4-9t^2+14&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(t^2-7)(t^2-2)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
t^2-7=0 & t^2-2=0
\\
t^2=7 & t^2=2
.\end{array}
Taking the square root of both sides (Square Root Property), the equations above are equivalent to
\begin{array}{l|r}
t=\pm\sqrt{7} & t=\pm\sqrt{2}
.\end{array}
Hence, the solution set of the equation $
t^4+14=9t^2
$ is $\left\{-\sqrt{7},-\sqrt{2},\sqrt{2},\sqrt{7}\right\}$.
