1. The problem is to find the root of $f(x) = x^2 - 3$ on the interval $[1, 2]$ using the bisection method with accuracy $\varepsilon = 0.01$. 2. Initial values are $a = 1$, $b = 2$. We compute $f(a) = 1^2 - 3 = -2$ and $f(b) = 2^2 - 3 = 1$. Since $f(a) \cdot f(b) < 0$, a root exists in $[a, b]$. 3. Compute midpoint $c_0 = \frac{1+2}{2} = 1.5$. Calculate $f(c_0) = 1.5^2 - 3 = 2.25 - 3 = -0.75$. 4. Check sign product $f(c_0) \cdot f(a) = (-0.75)(-2) = 1.5 > 0$, so update interval to $a_1 = 1.5$, $b_1 = 2$. 5. Compute midpoint $c_1 = \frac{1.5 + 2}{2} = 1.75$. Calculate $f(c_1) = 1.75^2 - 3 = 3.0625 - 3 = 0.0625$. 6. Check $|c_1 - c_0| = |1.75 - 1.5| = 0.25 > 0.01$, continue. 7. Check sign product $f(c_1) \cdot f(a_1) = (0.0625)(-0.75) = -0.046875 < 0$, so update interval to $a_2 = 1.5$, $b_2 = 1.75$. 8. Compute midpoint $c_2 = \frac{1.5 + 1.75}{2} = 1.625$. Calculate $f(c_2) = 1.625^2 - 3 = 2.640625 - 3 = -0.359375$. 9. Check $|c_2 - c_1| = |1.625 - 1.75| = 0.125 > 0.01$, continue. 10. Check sign product $f(c_2) \cdot f(a_2) = (-0.359375)(-0.75) = 0.26953125 > 0$, so update $a_3 = 1.625$, $b_3 = 1.75$. 11. Compute midpoint $c_3 = \frac{1.625 + 1.75}{2} = 1.6875$. Calculate $f(c_3) = 1.6875^2 - 3 = 2.84765625 - 3 = -0.15234375$. 12. Check $|c_3 - c_2| = |1.6875 - 1.625| = 0.0625 > 0.01$, continue. The iterative process continues until $|b - a| \leq \varepsilon$. The current approximation of the root is close to 1.6875. Final answer: The root of $f(x) = x^2 - 3$ on $[1,2]$ using the bisection method with accuracy 0.01 is approximately $x = 1.6875$.