{"id":428,"date":"2015-10-26T18:17:43","date_gmt":"2015-10-26T18:17:43","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=428"},"modified":"2015-11-12T18:37:59","modified_gmt":"2015-11-12T18:37:59","slug":"section-exercises-11","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/section-exercises-11\/","title":{"raw":"Section Exercises","rendered":"Section Exercises"},"content":{"raw":"1.\u00a0In a radical equation, what does it mean if a number is an extraneous solution?\r\n\r\n2.\u00a0Explain why possible solutions must<\/em> be checked in radical equations.\r\n\r\n3. Your friend tries to calculate the value [latex]-{9}^{\\frac{3}{2}}[\/latex] and keeps getting an ERROR message. What mistake is he or she probably making?\r\n\r\n4.\u00a0Explain why [latex]|2x+5|=-7[\/latex] has no solutions.\r\n\r\n5. Explain how to change a rational exponent into the correct radical expression.\r\n\r\nFor the following exercises, solve the rational exponent equation. Use factoring where necessary.\r\n\r\n6. [latex]{x}^{\\frac{2}{3}}=16[\/latex]\r\n\r\n7. [latex]{x}^{\\frac{3}{4}}=27[\/latex]\r\n\r\n8.\u00a0[latex]2{x}^{\\frac{1}{2}}-{x}^{\\frac{1}{4}}=0[\/latex]\r\n\r\n9. [latex]{\\left(x - 1\\right)}^{\\frac{3}{4}}=8[\/latex]\r\n\r\n10.\u00a0[latex]{\\left(x+1\\right)}^{\\frac{2}{3}}=4[\/latex]\r\n\r\n11. [latex]{x}^{\\frac{2}{3}}-5{x}^{\\frac{1}{3}}+6=0[\/latex]\r\n\r\n12.\u00a0[latex]{x}^{\\frac{7}{3}}-3{x}^{\\frac{4}{3}}-4{x}^{\\frac{1}{3}}=0[\/latex]\r\n\r\nFor the following exercises, solve the following polynomial equations by grouping and factoring.\r\n\r\n13. [latex]{x}^{3}+2{x}^{2}-x - 2=0[\/latex]\r\n\r\n14.\u00a0[latex]3{x}^{3}-6{x}^{2}-27x+54=0[\/latex]\r\n\r\n15. [latex]4{y}^{3}-9y=0[\/latex]\r\n\r\n16.\u00a0[latex]{x}^{3}+3{x}^{2}-25x - 75=0[\/latex]\r\n\r\n17. [latex]{m}^{3}+{m}^{2}-m - 1=0[\/latex]\r\n\r\n18.\u00a0[latex]2{x}^{5}-14{x}^{3}=0[\/latex]\r\n\r\n19. [latex]5{x}^{3}+45x=2{x}^{2}+18[\/latex]\r\n\r\nFor the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.\r\n\r\n20. [latex]\\sqrt{3x - 1}-2=0[\/latex]\r\n\r\n21. [latex]\\sqrt{x - 7}=5[\/latex]\r\n\r\n22.\u00a0[latex]\\sqrt{x - 1}=x - 7[\/latex]\r\n\r\n23. [latex]\\sqrt{3t+5}=7[\/latex]\r\n\r\n24.\u00a0[latex]\\sqrt{t+1}+9=7[\/latex]\r\n\r\n25. [latex]\\sqrt{12-x}=x[\/latex]\r\n\r\n26.\u00a0[latex]\\sqrt{2x+3}-\\sqrt{x+2}=2[\/latex]\r\n\r\n27. [latex]\\sqrt{3x+7}+\\sqrt{x+2}=1[\/latex]\r\n\r\n28.\u00a0[latex]\\sqrt{2x+3}-\\sqrt{x+1}=1[\/latex]\r\n\r\nFor the following exercises, solve the equation involving absolute value.\r\n\r\n29. [latex]|3x - 4|=8[\/latex]\r\n\r\n30.\u00a0[latex]|2x - 3|=-2[\/latex]\r\n\r\n31. [latex]|1 - 4x|-1=5[\/latex]\r\n\r\n32.\u00a0[latex]|4x+1|-3=6[\/latex]\r\n\r\n33. [latex]|2x - 1|-7=-2[\/latex]\r\n\r\n34.\u00a0[latex]|2x+1|-2=-3[\/latex]\r\n\r\n35. [latex]|x+5|=0[\/latex]\r\n\r\n36.\u00a0[latex]-|2x+1|=-3[\/latex]\r\n\r\nFor the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.\r\n\r\n37. [latex]{x}^{4}-10{x}^{2}+9=0[\/latex]\r\n\r\n38.\u00a0[latex]4{\\left(t - 1\\right)}^{2}-9\\left(t - 1\\right)=-2[\/latex]\r\n\r\n39. [latex]{\\left({x}^{2}-1\\right)}^{2}+\\left({x}^{2}-1\\right)-12=0[\/latex]\r\n\r\n40.\u00a0[latex]{\\left(x+1\\right)}^{2}-8\\left(x+1\\right)-9=0[\/latex]\r\n\r\n41. [latex]{\\left(x - 3\\right)}^{2}-4=0[\/latex]\r\n\r\nFor the following exercises, solve for the unknown variable.\r\n\r\n42. [latex]{x}^{-2}-{x}^{-1}-12=0[\/latex]\r\n\r\n43. [latex]\\sqrt{{|x|}^{2}}=x[\/latex]\r\n\r\n44.\u00a0[latex]{t}^{25}-{t}^{5}+1=0[\/latex]\r\n\r\n45. [latex]|{x}^{2}+2x - 36|=12[\/latex]\r\n\r\nFor the following exercises, use the model for the period of a pendulum, [latex]T[\/latex], such that [latex]T=2\\pi \\sqrt{\\frac{L}{g}}[\/latex], where the length of the pendulum is L<\/em> and the acceleration due to gravity is [latex]g[\/latex].\r\n\r\n46. If the acceleration due to gravity is [latex]9.8\\mathrm{m\/}{\\text{s}}^{2}[\/latex] and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).\r\n\r\n47. If the gravity is [latex]32\\frac{\\text{ft}}{{\\text{s}}^{2}}[\/latex] and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.\r\n\r\nFor the following exercises, use a model for body surface area, BSA, such that [latex]BSA=\\sqrt{\\frac{wh}{3600}}[\/latex], where w<\/em> = weight in kg and h<\/em> = height in cm.\r\n\r\n48. Find the height of a 72-kg female to the nearest cm whose [latex]BSA=1.8[\/latex].\r\n\r\n49. Find the weight of a 177-cm male to the nearest kg whose [latex]BSA=2.1[\/latex].","rendered":"

1.\u00a0In a radical equation, what does it mean if a number is an extraneous solution?<\/p>\n

2.\u00a0Explain why possible solutions must<\/em> be checked in radical equations.<\/p>\n

3. Your friend tries to calculate the value [latex]-{9}^{\\frac{3}{2}}[\/latex] and keeps getting an ERROR message. What mistake is he or she probably making?<\/p>\n

4.\u00a0Explain why [latex]|2x+5|=-7[\/latex] has no solutions.<\/p>\n

5. Explain how to change a rational exponent into the correct radical expression.<\/p>\n

For the following exercises, solve the rational exponent equation. Use factoring where necessary.<\/p>\n

6. [latex]{x}^{\\frac{2}{3}}=16[\/latex]<\/p>\n

7. [latex]{x}^{\\frac{3}{4}}=27[\/latex]<\/p>\n

8.\u00a0[latex]2{x}^{\\frac{1}{2}}-{x}^{\\frac{1}{4}}=0[\/latex]<\/p>\n

9. [latex]{\\left(x - 1\\right)}^{\\frac{3}{4}}=8[\/latex]<\/p>\n

10.\u00a0[latex]{\\left(x+1\\right)}^{\\frac{2}{3}}=4[\/latex]<\/p>\n

11. [latex]{x}^{\\frac{2}{3}}-5{x}^{\\frac{1}{3}}+6=0[\/latex]<\/p>\n

12.\u00a0[latex]{x}^{\\frac{7}{3}}-3{x}^{\\frac{4}{3}}-4{x}^{\\frac{1}{3}}=0[\/latex]<\/p>\n

For the following exercises, solve the following polynomial equations by grouping and factoring.<\/p>\n

13. [latex]{x}^{3}+2{x}^{2}-x - 2=0[\/latex]<\/p>\n

14.\u00a0[latex]3{x}^{3}-6{x}^{2}-27x+54=0[\/latex]<\/p>\n

15. [latex]4{y}^{3}-9y=0[\/latex]<\/p>\n

16.\u00a0[latex]{x}^{3}+3{x}^{2}-25x - 75=0[\/latex]<\/p>\n

17. [latex]{m}^{3}+{m}^{2}-m - 1=0[\/latex]<\/p>\n

18.\u00a0[latex]2{x}^{5}-14{x}^{3}=0[\/latex]<\/p>\n

19. [latex]5{x}^{3}+45x=2{x}^{2}+18[\/latex]<\/p>\n

For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions.<\/p>\n

20. [latex]\\sqrt{3x - 1}-2=0[\/latex]<\/p>\n

21. [latex]\\sqrt{x - 7}=5[\/latex]<\/p>\n

22.\u00a0[latex]\\sqrt{x - 1}=x - 7[\/latex]<\/p>\n

23. [latex]\\sqrt{3t+5}=7[\/latex]<\/p>\n

24.\u00a0[latex]\\sqrt{t+1}+9=7[\/latex]<\/p>\n

25. [latex]\\sqrt{12-x}=x[\/latex]<\/p>\n

26.\u00a0[latex]\\sqrt{2x+3}-\\sqrt{x+2}=2[\/latex]<\/p>\n

27. [latex]\\sqrt{3x+7}+\\sqrt{x+2}=1[\/latex]<\/p>\n

28.\u00a0[latex]\\sqrt{2x+3}-\\sqrt{x+1}=1[\/latex]<\/p>\n

For the following exercises, solve the equation involving absolute value.<\/p>\n

29. [latex]|3x - 4|=8[\/latex]<\/p>\n

30.\u00a0[latex]|2x - 3|=-2[\/latex]<\/p>\n

31. [latex]|1 - 4x|-1=5[\/latex]<\/p>\n

32.\u00a0[latex]|4x+1|-3=6[\/latex]<\/p>\n

33. [latex]|2x - 1|-7=-2[\/latex]<\/p>\n

34.\u00a0[latex]|2x+1|-2=-3[\/latex]<\/p>\n

35. [latex]|x+5|=0[\/latex]<\/p>\n

36.\u00a0[latex]-|2x+1|=-3[\/latex]<\/p>\n

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring.<\/p>\n

37. [latex]{x}^{4}-10{x}^{2}+9=0[\/latex]<\/p>\n

38.\u00a0[latex]4{\\left(t - 1\\right)}^{2}-9\\left(t - 1\\right)=-2[\/latex]<\/p>\n

39. [latex]{\\left({x}^{2}-1\\right)}^{2}+\\left({x}^{2}-1\\right)-12=0[\/latex]<\/p>\n

40.\u00a0[latex]{\\left(x+1\\right)}^{2}-8\\left(x+1\\right)-9=0[\/latex]<\/p>\n

41. [latex]{\\left(x - 3\\right)}^{2}-4=0[\/latex]<\/p>\n

For the following exercises, solve for the unknown variable.<\/p>\n

42. [latex]{x}^{-2}-{x}^{-1}-12=0[\/latex]<\/p>\n

43. [latex]\\sqrt{{|x|}^{2}}=x[\/latex]<\/p>\n

44.\u00a0[latex]{t}^{25}-{t}^{5}+1=0[\/latex]<\/p>\n

45. [latex]|{x}^{2}+2x - 36|=12[\/latex]<\/p>\n

For the following exercises, use the model for the period of a pendulum, [latex]T[\/latex], such that [latex]T=2\\pi \\sqrt{\\frac{L}{g}}[\/latex], where the length of the pendulum is L<\/em> and the acceleration due to gravity is [latex]g[\/latex].<\/p>\n

46. If the acceleration due to gravity is [latex]9.8\\mathrm{m\/}{\\text{s}}^{2}[\/latex] and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m).<\/p>\n

47. If the gravity is [latex]32\\frac{\\text{ft}}{{\\text{s}}^{2}}[\/latex] and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in.<\/p>\n

For the following exercises, use a model for body surface area, BSA, such that [latex]BSA=\\sqrt{\\frac{wh}{3600}}[\/latex], where w<\/em> = weight in kg and h<\/em> = height in cm.<\/p>\n

48. Find the height of a 72-kg female to the nearest cm whose [latex]BSA=1.8[\/latex].<\/p>\n

49. Find the weight of a 177-cm male to the nearest kg whose [latex]BSA=2.1[\/latex].<\/p>\n\n\t\t\t

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