1. **State the problem:** We have a quadratic equation $$ax^2 + b = 181$$ where $$a$$ and $$b$$ are positive integers greater than 1. One solution is $$x=8$$. We need to find (a) the value of $$b$$ and (b) the other solution. 2. **Use the given solution to find a relation:** Substitute $$x=8$$ into the equation: $$a \times 8^2 + b = 181$$ which simplifies to $$64a + b = 181$$ 3. **Find possible values of $$a$$ and $$b$$:** Since $$a$$ and $$b$$ are positive integers greater than 1, rewrite: $$b = 181 - 64a$$ 4. **Check integer values of $$a > 1$$ to find integer $$b > 1$$:** - For $$a=2$$: $$b = 181 - 128 = 53$$ (valid) - For $$a=3$$: $$b = 181 - 192 = -11$$ (invalid, negative) So the only valid pair is $$a=2$$ and $$b=53$$. 5. **Find the other solution:** The quadratic equation is $$2x^2 + 53 = 181$$ which simplifies to $$2x^2 = 128$$ $$x^2 = 64$$ 6. **Solve for $$x$$:** $$x = \pm \sqrt{64} = \pm 8$$ Since one solution is $$x=8$$, the other solution is $$x=-8$$. **Final answers:** (a) $$b = 53$$ (b) $$x = -8$$