Answer
$x=\dfrac{3-\sqrt{5}}{2}
\text{ and }
x=\dfrac{3+\sqrt{5}}{2}
$
Work Step by Step
Using the properties of equality, the given equation, $
x^2=3x-1
,$ is equivalent to
\begin{align*}
x^2-3x+1=0
\end{align*}
In the equation above, $a=
1
,$ $b=
-3
,$ and $c=
1
.$
Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then
\begin{align*}x&=
\dfrac{-(-3)\pm\sqrt{(-3)^2-4(1)(1)}}{2(1)}
\\\\&=
\dfrac{3\pm\sqrt{9-4}}{2}
\\\\&=
\dfrac{3\pm\sqrt{5}}{2}
\end{align*}
Hence, the solutions are $
x=\dfrac{3-\sqrt{5}}{2}
\text{ and }
x=\dfrac{3+\sqrt{5}}{2}
.$
