**TEST ONE: Algebra Homework Assignment** 1. Solve the equation $\sqrt{x-8} - \sqrt{2x-2} + 3 = 0$. 1. Start by isolating one square root: $$\sqrt{x-8} = \sqrt{2x-2} - 3$$ 2. Square both sides to eliminate the square root on the left: $$x - 8 = (\sqrt{2x-2} - 3)^2$$ 3. Expand the right side: $$x - 8 = 2x - 2 - 6\sqrt{2x-2} + 9$$ 4. Rearrange terms: $$-x - 15 = -6 \sqrt{2x-2}$$ 5. Divide both sides by $-6$: $$\frac{x+15}{6} = \sqrt{2x-2}$$ 6. Square both sides again: $$\left(\frac{x+15}{6}\right)^2 = 2x - 2$$ 7. Simplify: $$\frac{(x+15)^2}{36} = 2x - 2$$ 8. Multiply both sides by 36: $$ (x+15)^2 = 72x - 72$$ 9. Expand left side: $$x^2 + 30x + 225 = 72x - 72$$ 10. Bring all terms to one side: $$x^2 + 30x + 225 - 72x + 72 = 0$$ $$x^2 - 42x + 297 = 0$$ 11. Solve quadratic equation using the quadratic formula: $$x = \frac{42 \pm \sqrt{(-42)^2 - 4 \times 1 \times 297}}{2} = \frac{42 \pm \sqrt{1764 - 1188}}{2} = \frac{42 \pm \sqrt{576}}{2}$$ $$x = \frac{42 \pm 24}{2}$$ 12. Possible solutions: $$x = \frac{42 + 24}{2} = 33, \quad x = \frac{42 - 24}{2} = 9$$ 13. Check solutions in original equation: - For $x=33$: $$\sqrt{33-8} - \sqrt{2(33)-2} + 3 = \sqrt{25} - \sqrt{64} + 3 = 5 - 8 + 3 = 0$$ (valid) - For $x=9$: $$\sqrt{9-8} - \sqrt{18-2} + 3 = \sqrt{1} - \sqrt{16} + 3 = 1 - 4 + 3 = 0$$ (valid) 14. Final solutions: $$\boxed{x = 9 \text{ or } x = 33}$$ 2. Solve for $x$ in the equation: $$\frac{1-\sqrt{2}}{2-\sqrt{2}} = \frac{\sqrt{3}x}{3\sqrt{3}}$$ 1. Simplify right side: $$\frac{\sqrt{3}x}{3\sqrt{3}} = \frac{x}{3}$$ 2. Rationalize the left side denominator: $$\frac{1-\sqrt{2}}{2-\sqrt{2}} \times \frac{2+\sqrt{2}}{2+\sqrt{2}} = \frac{(1-\sqrt{2})(2+\sqrt{2})}{(2)^2 - (\sqrt{2})^2} = \frac{(1-\sqrt{2})(2+\sqrt{2})}{4 - 2} = \frac{(1-\sqrt{2})(2+\sqrt{2})}{2}$$ 3. Expand numerator: $$2 + \sqrt{2} - 2\sqrt{2} - 2 = 0 - \sqrt{2} = -\sqrt{2}$$ 4. So left side is: $$\frac{-\sqrt{2}}{2}$$ 5. Set equal to right side: $$\frac{-\sqrt{2}}{2} = \frac{x}{3}$$ 6. Cross multiply: $$x = -\frac{3\sqrt{2}}{2}$$ 7. Final answer: $$\boxed{x = -\frac{3\sqrt{2}}{2}}$$ 3. Find the positive square root of $7 - \sqrt{40}$. 1. Simplify $\sqrt{40}$: $$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$$ 2. So expression is: $$7 - 2\sqrt{10}$$ 3. Assume positive root is of form $\sqrt{a} - \sqrt{b}$: $$\left(\sqrt{a} - \sqrt{b}\right)^2 = a + b - 2\sqrt{ab}$$ 4. Match terms: $$a + b = 7, \quad 2\sqrt{ab} = 2\sqrt{10} \Rightarrow \sqrt{ab} = \sqrt{10} \Rightarrow ab = 10$$ 5. Solve system: $$a + b = 7, \quad ab = 10$$ 6. Possible $a,b$ are 5 and 2. 7. Therefore: $$\sqrt{7 - 2\sqrt{10}} = \sqrt{5} - \sqrt{2}$$ 8. Final answer: $$\boxed{\sqrt{5} - \sqrt{2}}$$ 4. Given: $$\frac{\sqrt{5}-4}{3+2\sqrt{5}} - \frac{2-\sqrt{5}}{4+2\sqrt{5}} = m + n\sqrt{5}$$ Find the product $mn$. 1. Rationalize denominators: $$\frac{\sqrt{5}-4}{3+2\sqrt{5}} \times \frac{3-2\sqrt{5}}{3-2\sqrt{5}} = \frac{(\sqrt{5}-4)(3-2\sqrt{5})}{(3)^2 - (2\sqrt{5})^2} = \frac{(\sqrt{5}-4)(3-2\sqrt{5})}{9 - 20} = \frac{(\sqrt{5}-4)(3-2\sqrt{5})}{-11}$$ 2. Similarly for second term: $$\frac{2-\sqrt{5}}{4+2\sqrt{5}} \times \frac{4-2\sqrt{5}}{4-2\sqrt{5}} = \frac{(2-\sqrt{5})(4-2\sqrt{5})}{16 - 20} = \frac{(2-\sqrt{5})(4-2\sqrt{5})}{-4}$$ 3. Compute numerators: - First numerator: $$(\sqrt{5}-4)(3-2\sqrt{5}) = 3\sqrt{5} - 6\sqrt{5} - 12 + 8\sqrt{5} = (3\sqrt{5} - 8\sqrt{5} + 8\sqrt{5}) - 12 = 3\sqrt{5} - 12$$ (Recalculate carefully: actually, $$3\sqrt{5} - 2\sqrt{5} \times \sqrt{5} - 12 + 8\sqrt{5} = 3\sqrt{5} - 2 \times 5 - 12 + 8\sqrt{5} = 3\sqrt{5} - 10 - 12 + 8\sqrt{5} = (3\sqrt{5} + 8\sqrt{5}) - 22 = 11\sqrt{5} - 22$$) 4. So first term: $$\frac{11\sqrt{5} - 22}{-11} = -\sqrt{5} + 2$$ 5. Second numerator: $$(2-\sqrt{5})(4-2\sqrt{5}) = 8 - 4\sqrt{5} - 4\sqrt{5} + 2 \times 5 = 8 - 8\sqrt{5} + 10 = 18 - 8\sqrt{5}$$ 6. Second term: $$\frac{18 - 8\sqrt{5}}{-4} = -\frac{18}{4} + 2\sqrt{5} = -\frac{9}{2} + 2\sqrt{5}$$ 7. Combine terms: $$(-\sqrt{5} + 2) - \left(-\frac{9}{2} + 2\sqrt{5}\right) = -\sqrt{5} + 2 + \frac{9}{2} - 2\sqrt{5} = \left(2 + \frac{9}{2}\right) + (-\sqrt{5} - 2\sqrt{5}) = \frac{13}{2} - 3\sqrt{5}$$ 8. So: $$m = \frac{13}{2}, \quad n = -3$$ 9. Product: $$mn = \frac{13}{2} \times (-3) = -\frac{39}{2}$$ 10. Final answer: $$\boxed{-\frac{39}{2}}$$ **Summary:** - Problem 1 solutions: $x=9$ or $x=33$ - Problem 2 solution: $x = -\frac{3\sqrt{2}}{2}$ - Problem 3 positive root: $\sqrt{5} - \sqrt{2}$ - Problem 4 product $mn = -\frac{39}{2}$