Answer
$1.354977808$
Work Step by Step
Step 1. Take the difference of the two functions and let $f(x)=x^2+\sqrt x-3, x\gt0$; we have $f'(x)=2x+\frac{1}{2\sqrt x}$. Finding the intersection of the graphs given in the exercise is the same as finding the zero of $f(x)$.
Step 2. Test signs of $f(x)$, $f(1)=1+1-3=-1\lt0$ and $f(4)=16+2-3=15\gt0$; thus there is a zero in the interval of $(1,4)$
Step 3. Using Newton’s method $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ with a starting point $x_0=1$, we can obtain the zero shown in the table as $1.354977808$
