(d) \( \sqrt{8-x}=\sqrt{x-2} \) 2 Find the values of the integers \( a \) and \( b \). (a) \( a+b \sqrt{3}=2-\sqrt{3} \) (b) \( a+b \sqrt{7}=4 \sqrt{7}-12-\sqrt{7} \) (c) \( a+b \sqrt{2}=3(1-\sqrt{2})+4 \sqrt{2} \)

We have two separate problems. ───────────────────────────── Problem (d): Solve √(8 – x) = √(x – 2). Step 1. Note the domains: • For √(8 – x) to be defined, 8 – x ≥ 0 ⇒ x ≤ 8. • For √(x – 2) to be defined, x – 2 ≥ 0 ⇒ x ≥ 2. Thus, x must lie in the interval [2, 8]. Step 2. Since both square roots are nonnegative, we can square each side of the equation: (√(8 – x))² = (√(x – 2))² ⇒ 8 – x = x – 2. Step 3. Solve for x: 8 – x = x – 2 ⇒ 8 + 2 = x + x ⇒ 10 = 2x ⇒ x = 5. Step 4. Check that x = 5 satisfies the domain: (5 is between 2 and 8.) Thus, x = 5 is the solution. ───────────────────────────── Problem 2: Find the integers a and b in each case. (a) a + b√3 = 2 – √3. Here, equate the rational and irrational parts: • Rational: a = 2 • Irrational: b√3 = –√3, which implies b = –1. (b) a + b√7 = 4√7 – 12 – √7. First, simplify the right-hand side: 4√7 – √7 = 3√7, so we have: a + b√7 = 3√7 – 12. Now equate the components: • Rational: a = –12 • Irrational: b√7 = 3√7, so b = 3. (c) a + b√2 = 3(1 – √2) + 4√2. Expand the expression: 3(1 – √2) = 3 – 3√2. Thus, 3 – 3√2 + 4√2 = 3 + (–3√2 + 4√2) = 3 + √2. So we have: a + b√2 = 3 + √2. Equate the parts: • Rational: a = 3 • Irrational: b√2 = √2, which means b = 1. ───────────────────────────── Final Answers: For (d), the solution is x = 5. For Problem 2: (a) a = 2 and b = –1; (b) a = –12 and b = 3; (c) a = 3 and b = 1.