Question

the square root of 3t+1 = the square root of t+9

Answer 1

\[\sqrt{3t+1} = \sqrt{t+9}\]

Are we doing this problem?

Answer 2

Solving for t I'm guessing.

Answer 3

\[3 t + 1 = t + 9 \]
t = 4

Answer 4

Square each side.

Answer 5

Similar example,
\(\color{#000000}{ \displaystyle \sqrt{2x+5}=\sqrt{5x-1} }\)

Take the square on both sides
(raise both sides to the second power)
\(\color{#000000}{ \displaystyle \color{red}{\left(\color{black}{\sqrt{2x+5}}\right)^2} =\color{red}{\left(\color{black}{\sqrt{5x-1}}\right)^2} }\)

There is a rule,
\(\color{#000000}{ \displaystyle (\sqrt[n]{z})^n=z }\)
(and same way, \(\color{#000000}{ \displaystyle (\sqrt[2]{z})^2=z }\))
so, in this case you get,
\(\color{#000000}{ \displaystyle 2x+5=5x-1 }\)

From now you can solve this arithmetically.
\(\color{#000000}{ \displaystyle 2x+5\color{red}{-2x}=5x-1\color{red}{-2x} }\)
\(\color{#000000}{ \displaystyle 5=3x-1 }\)
\(\color{#000000}{ \displaystyle 5\color{red}{+1}=3x-1\color{red}{+1} }\)
\(\color{#000000}{ \displaystyle 6=3x }\)
\(\color{#000000}{ \displaystyle 6\color{red}{\div 3}=3x\color{red}{\div 3} }\)
\(\color{#000000}{ \displaystyle 2=x }\)

If you want/need to check,
\(\color{#000000}{ \displaystyle \sqrt{2x+5}=\sqrt{5x-1} }\)
\(\color{#000000}{ \displaystyle \sqrt{2\color{red}{(2)} +5}=\sqrt{5\color{red}{(2)}-1} }\)
\(\color{#000000}{ \displaystyle \sqrt{4+5}=\sqrt{10-1} }\)
\(\color{#000000}{ \displaystyle \sqrt{9}=\sqrt{9} }\)

Answer 6

\[\sqrt{3t+1} = \sqrt{t+9}\]