Answer
No solution
Work Step by Step
Given \begin{equation}
\sqrt{2 x+3}=1-\sqrt{x+5}.
\end{equation} First square both sides of the radical equation to eliminate the radical sign on the left. Rearrange thequation and square both sides again to eliminate the radical sign on the right hand side and then solve for $x$.
\begin{equation}
\begin{aligned}
\sqrt{2 x+3}&=1-\sqrt{x+5}\\
\left( \sqrt{2 x+3}\right)^2&= \left( 1-\sqrt{x+5}\right)^2\\
2 x+3& = 1-2\sqrt{x+5}+x+5\\
2x-x+3-6& = -2\sqrt{x+5}\\
x-3 & = -2\sqrt{x+5}\\
\left(x-3\right)^2& = \left(-2\sqrt{x+5}\right)^2\\
x^2-6x+9&= 4x+20\\
x^2-10x-11&=0.
\end{aligned}
\end{equation} Use the quadratic formula to find the value(s) of $x$.
\begin{equation}
\begin{aligned}
a&x^2+bx+c=0\\
a &= 1\ , b= -10\ ,\ c = -11\\
x & = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\
x&=\frac{-(-10) \pm \sqrt{(-10)^2-4 \cdot 1 \cdot(-11)}}{2 \cdot 1} \\
&= \frac{10\pm 12}{2}\\
\implies x& = \frac{10- 12}{2}\\
& = -1\\
x& = \frac{10+12}{2}\\
& = 11.
\end{aligned}
\end{equation} Check \begin{equation}
\begin{aligned}
\sqrt{2\cdot (-1)+3}& \stackrel{?}{=} 1-\sqrt{-1+5} \\
1& \stackrel{?}{=}-1\quad \textbf{False}\\
\sqrt{2\cdot (11)+3}& \stackrel{?}{=} 1-\sqrt{11+5} \\
5& \stackrel{?}{=}-3\quad \textbf{False}\
\end{aligned}
\end{equation} The equation has no real solution.
