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Use the square root property to solve the equation. t^2=98 please use this formular -If t is a positive #, and if x^2=t, then x=sqrt t or x=-qurt t
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t=sqrt 98=9.89
p/s use this formular if b is a positive # and x^2=k then x=sqrt K or x=-sqrt k
\[t^2=98\] \[t=\pm\sqrt{98}\]
since \[98=2\times 49\] you can write \[t=\pm7\sqrt{2}\] if you like
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