Answer
$x=\{1,9\}$
Work Step by Step
Squaring both sides of the given equation, $
\sqrt{16x}=x+3
,$ then,
\begin{array}{l}\require{cancel}
\left( \sqrt{16x} \right)^2=\left( x+3 \right)^2
\\\\
16x=(x)^2+2(x)(3)+(3)^2
\\\\
16x=x^2+6x+9
\\\\
-x^2+16x-6x-9=0
\\\\
-x^2+10x-9=0
.\end{array}
Using $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ (or the Quadratic Formula), the solutions of the quadratic equation, $
-x^2+10x-9=0
,$ are
\begin{array}{l}\require{cancel}
\dfrac{-(10)\pm\sqrt{(10)^2-4(-1)(-9)}}{2(-1)}
\\\\=
\dfrac{-10\pm\sqrt{100-36}}{-2}
\\\\=
\dfrac{-10\pm\sqrt{64}}{-2}
\\\\=
\dfrac{-10\pm8}{-2}
\\\\=
\dfrac{-10-8}{-2}
\text{ OR }
\dfrac{-10+8}{-2}
\\\\=
\dfrac{-18}{-2}
\text{ OR }
\dfrac{-2}{-2}
\\\\=
9
\text{ OR }
1
.\end{array}
Upon checking, both solutions satisfy the original equation. Hence, $
x=\{1,9\}
.$
