College Algebra, Chapter 2, 2.3, Section 2.3, Problem 48

Solve the equation $1 + \sqrt{x} = \sqrt{1+x^2}$ graphically on the interval $[-1,5]$. State each answer correct to two decimals.



Based from the graph, the equation $1 + \sqrt{x}$ and $\sqrt{1+x^2}$ is equal when $x \approx 0$ and $x \approx 2.20$
By solving the exact value,

$\begin{equation} \begin{aligned} 1 + \sqrt{x} &= \sqrt{1+x^2} \\ \\ 1 + 2\sqrt{x} + x &= 1 + x^2 && \text{Square both sides}\\ \\ 2\sqrt{x} + x &= x^2 && \text{Subtract } 1\\ \\ 2\sqrt{x} &= x^2 - x && \text{Subtract } x \\ \\ 4x &= x^4 - 2x^3 + x^2 && \text{Square both sides}\\ \\ x^4 - 2x^3 + x^2 - 4x &= 0 && \text{Subtract } 4x\\ \\ x\left( x^3 - 2x^2 + x - 4 \right) &= 0 && \text{Factor out } x\\ \\ x = 0 \text{ and }x^3 - 2x^2 + x - 4 &= 0 && \text{Zero product property} \end{aligned} \end{equation}$


By using calculator the solutions $x^3 - 2x^2 + x - 4 = 0$ are $x = 2.3146, \quad x = 0.1572+1.305i, \quad$ and $x = -0.1572 - 1.305i$
Thus, the real solution for the equation is...
$x = 0$ and $x = 2.3146$