1. **State the problem:** Two resistors $a$ and $b$ are connected in parallel. The total resistance $R$ is given by the formula $$R = \frac{1}{\frac{1}{a} + \frac{1}{b}}.$$ We know $R = \frac{5}{6}$ ohms and $b = a + 1$. We need to find the values of $a$ and $b$ where both are positive. 2. **Write the formula for total resistance in parallel:** $$R = \frac{1}{\frac{1}{a} + \frac{1}{b}} = \frac{1}{\frac{b+a}{ab}} = \frac{ab}{a+b}.$$ 3. **Substitute $R = \frac{5}{6}$ and $b = a + 1$ into the formula:** $$\frac{5}{6} = \frac{a(a+1)}{a + (a+1)} = \frac{a(a+1)}{2a + 1}.$$ 4. **Cross multiply to clear the fraction:** $$5(2a + 1) = 6a(a+1).$$ 5. **Expand both sides:** $$10a + 5 = 6a^2 + 6a.$$ 6. **Bring all terms to one side to form a quadratic equation:** $$6a^2 + 6a - 10a - 5 = 0 \implies 6a^2 - 4a - 5 = 0.$$ 7. **Solve the quadratic equation $6a^2 - 4a - 5 = 0$ using the quadratic formula:** $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \times 6 \times (-5)}}{2 \times 6} = \frac{4 \pm \sqrt{16 + 120}}{12} = \frac{4 \pm \sqrt{136}}{12}.$$ 8. **Simplify $\sqrt{136}$:** $$\sqrt{136} = \sqrt{4 \times 34} = 2\sqrt{34}.$$ 9. **So,** $$a = \frac{4 \pm 2\sqrt{34}}{12} = \frac{2 \pm \sqrt{34}}{6}.$$ 10. **Since $a > 0$, choose the positive root:** $$a = \frac{2 + \sqrt{34}}{6}.$$ 11. **Calculate $b = a + 1$:** $$b = \frac{2 + \sqrt{34}}{6} + 1 = \frac{2 + \sqrt{34} + 6}{6} = \frac{8 + \sqrt{34}}{6}.$$ **Final answer:** $$a = \frac{2 + \sqrt{34}}{6} \approx 1.457, \quad b = \frac{8 + \sqrt{34}}{6} \approx 2.457.$$