1. **State the problem:** We want to compute the square root of 3 using the bisection method with a tolerance of $10^{-2}$. The bisection method finds roots of a function by repeatedly halving an interval where the function changes sign. 2. **Define the function:** Let $f(x) = x^2 - 3$. We want to find $x$ such that $f(x) = 0$, which corresponds to $x = \sqrt{3}$. 3. **Choose initial interval:** Since $1^2 = 1 < 3$ and $2^2 = 4 > 3$, the root lies in the interval $[1, 2]$. 4. **Bisection steps:** - Compute midpoint $m = \frac{a+b}{2}$. - Evaluate $f(m)$. - If $f(m) = 0$ or the interval length $|b - a| < 10^{-2}$, stop. - Otherwise, replace $a$ or $b$ with $m$ depending on the sign of $f(m)$. 5. **MATLAB code snippet:** ```matlab f = @(x) x.^2 - 3; a = 1; b = 2; tol = 1e-2; while (b - a) / 2 > tol m = (a + b) / 2; if f(m) == 0 break; elseif f(a) * f(m) < 0 b = m; else a = m; end end root = (a + b) / 2; disp(root) ``` 6. **Explanation:** The code repeatedly halves the interval $[a,b]$ until the interval length is less than the tolerance $10^{-2}$. The midpoint at the end is an approximation of $\sqrt{3}$. 7. **Final answer:** The approximate square root of 3 with tolerance $10^{-2}$ is about $1.73$.