Answer
$x=2$
Work Step by Step
Squaring both sides of the given equation, $
3\sqrt{x-1}=5-x
,$ results to
\begin{align*}\require{cancel}
\left(3\sqrt{x-1}\right)^2&=(5-x)^2
\\
9(x-1)&=(5-x)^2
\\
9(x-1)&=(5)^2-2(5)(x)+(-x)^2
&(\text{use }(a-b)^2=a^2-2ab+b^2)
\\
9(x-1)&=25-10x+x^2
\\
9(x)+9(-1)&=25-10x+x^2
&(\text{use the Distributive Property)}
\\
9x-9&=25-10x+x^2
\\
0&=(25+9)+(-10x-9x)+x^2
\\
0&=34-19x+x^2
\\
x^2-19x+34&=0
.\end{align*}
Using the factoring of trinomials, the equation above is equivalent to
\begin{align*}
(x-17)(x-2)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable results to
\begin{align*}
\begin{array}{l|r}
x-17=0 & x-2=0
\\
x-17+17=0+17 & x-2+2=0+2
\\
x=17 & x=2.
\end{array}
\end{align*}
Since both sides of the given radical equation were squared, the solution must be checked:
\begin{align*}
\begin{array}{l|r}
\text{If }x=17: & \text{If }x=2:
\\\\
3\sqrt{17-1}\overset{?}=5-17 & 3\sqrt{2-1}\overset{?}=5-2
\\
3\sqrt{16}\overset{?}=-12 & 3\sqrt{1}\overset{?}=3
\\
3(4)\overset{?}=-12 & 3(1)\overset{?}=3
\\
12\ne-12 & 3\overset{\checkmark}=3
\end{array}
\end{align*}
Hence, the only solution to $3\sqrt{x-1}=5-x$ is $x=2$.
