{"id":408,"date":"2015-10-26T17:58:56","date_gmt":"2015-10-26T17:58:56","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=408"},"modified":"2015-11-12T18:38:00","modified_gmt":"2015-11-12T18:38:00","slug":"solutions-to-selected-exercises-2","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/solutions-to-selected-exercises-2\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"

Solutions to Try Its<\/h2>\r\n1.\u00a0[latex]\\left(x - 6\\right)\\left(x+1\\right)=0;x=6,x=-1[\/latex]\r\n\r\n2.\u00a0[latex]\\left(x - 7\\right)\\left(x+3\\right)=0[\/latex], [latex]x=7[\/latex], [latex]x=-3[\/latex].\r\n\r\n3.\u00a0[latex]\\left(x+5\\right)\\left(x - 5\\right)=0[\/latex], [latex]x=-5[\/latex], [latex]x=5[\/latex].\r\n\r\n4.\u00a0[latex]\\left(3x+2\\right)\\left(4x+1\\right)=0[\/latex], [latex]x=-\\frac{2}{3}[\/latex], [latex]x=-\\frac{1}{4}[\/latex]\r\n\r\n5.\u00a0[latex]x=0,x=-10,x=-1[\/latex]\r\n\r\n6.\u00a0[latex]x=4\\pm \\sqrt{5}[\/latex]\r\n\r\n7.\u00a0[latex]x=3\\pm \\sqrt{22}[\/latex]\r\n\r\n8.\u00a0[latex]x=-\\frac{2}{3}[\/latex], [latex]x=\\frac{1}{3}[\/latex]\r\n\r\n9.\u00a0[latex]5[\/latex] units\r\n

Solutions to Odd-Numbered Exercises<\/h2>\r\n1.\u00a0It is a second-degree equation (the highest variable exponent is 2).\r\n\r\n3.\u00a0We want to take advantage of the zero property of multiplication in the fact that if [latex]a\\cdot b=0[\/latex] then it must follow that each factor separately offers a solution to the product being zero: [latex]a=0\\text{ }or\\text{ b}=0[\/latex].\r\n\r\n5.\u00a0One, when no linear term is present (no x<\/em> term), such as [latex]{x}^{2}=16[\/latex]. Two, when the equation is already in the form [latex]{\\left(ax+b\\right)}^{2}=d[\/latex].\r\n\r\n7.\u00a0[latex]x=6[\/latex], [latex]x=3[\/latex]\r\n\r\n9.\u00a0[latex]x=\\frac{-5}{2}[\/latex], [latex]x=\\frac{-1}{3}[\/latex]\r\n\r\n11.\u00a0[latex]x=5[\/latex], [latex]x=-5[\/latex]\r\n\r\n13.\u00a0[latex]x=\\frac{-3}{2}[\/latex], [latex]x=\\frac{3}{2}[\/latex]\r\n\r\n15.\u00a0[latex]x=-2[\/latex]\r\n\r\n17.\u00a0[latex]x=0[\/latex], [latex]x=\\frac{-3}{7}[\/latex]\r\n\r\n19.\u00a0[latex]x=-6[\/latex], [latex]x=6[\/latex]\r\n\r\n21.\u00a0[latex]x=6[\/latex], [latex]x=-4[\/latex]\r\n\r\n23.\u00a0[latex]x=1[\/latex], [latex]x=-2[\/latex]\r\n\r\n25.\u00a0[latex]x=-2[\/latex], [latex]x=11[\/latex]\r\n\r\n27.\u00a0[latex]x=3\\pm \\sqrt{22}[\/latex]\r\n\r\n29.\u00a0[latex]z=\\frac{2}{3}\\\\[\/latex], [latex]z=-\\frac{1}{2}[\/latex]\r\n\r\n31.\u00a0[latex]x=\\frac{3\\pm \\sqrt{17}}{4}[\/latex]\r\n\r\n33.\u00a0Not real\r\n\r\n35.\u00a0One rational\r\n\r\n37.\u00a0Two real; rational\r\n\r\n39.\u00a0[latex]x=\\frac{-1\\pm \\sqrt{17}}{2}[\/latex]\r\n\r\n41.\u00a0[latex]x=\\frac{5\\pm \\sqrt{13}}{6}[\/latex]\r\n\r\n43.\u00a0[latex]x=\\frac{-1\\pm \\sqrt{17}}{8}[\/latex]\r\n\r\n45.\u00a0[latex]x\\approx 0.131[\/latex] and [latex]x\\approx 2.535[\/latex]\r\n\r\n47.\u00a0[latex]x\\approx -6.7[\/latex] and [latex]x\\approx 1.7[\/latex]\r\n\r\n49.\u00a0[latex]\\begin{array}{l}a{x}^{2}+bx+c \\hfill& =0\\hfill \\\\ {x}^{2}+\\frac{b}{a}x \\hfill& =\\frac{-c}{a}\\hfill \\\\ {x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}} \\hfill& =\\frac{-c}{a}+\\frac{b}{4{a}^{2}}\\hfill \\\\ {\\left(x+\\frac{b}{2a}\\right)}^{2}\\hfill& =\\frac{{b}^{2}-4ac}{4{a}^{2}}\\hfill \\\\ x+\\frac{b}{2a}\\hfill& =\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x\\hfill& =\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]\r\n\r\n51.\u00a0[latex]x\\left(x+10\\right)=119[\/latex]; 7 ft. and 17 ft.\r\n\r\n53.\u00a0maximum at [latex]x=70[\/latex]\r\n\r\n55.\u00a0The quadratic equation would be [latex]\\left(100x - 0.5{x}^{2}\\right)-\\left(60x+300\\right)=300[\/latex]. The two values of [latex]x[\/latex] are 20 and 60.\r\n\r\n57.\u00a03 feet","rendered":"

Solutions to Try Its<\/h2>\n

1.\u00a0[latex]\\left(x - 6\\right)\\left(x+1\\right)=0;x=6,x=-1[\/latex]<\/p>\n

2.\u00a0[latex]\\left(x - 7\\right)\\left(x+3\\right)=0[\/latex], [latex]x=7[\/latex], [latex]x=-3[\/latex].<\/p>\n

3.\u00a0[latex]\\left(x+5\\right)\\left(x - 5\\right)=0[\/latex], [latex]x=-5[\/latex], [latex]x=5[\/latex].<\/p>\n

4.\u00a0[latex]\\left(3x+2\\right)\\left(4x+1\\right)=0[\/latex], [latex]x=-\\frac{2}{3}[\/latex], [latex]x=-\\frac{1}{4}[\/latex]<\/p>\n

5.\u00a0[latex]x=0,x=-10,x=-1[\/latex]<\/p>\n

6.\u00a0[latex]x=4\\pm \\sqrt{5}[\/latex]<\/p>\n

7.\u00a0[latex]x=3\\pm \\sqrt{22}[\/latex]<\/p>\n

8.\u00a0[latex]x=-\\frac{2}{3}[\/latex], [latex]x=\\frac{1}{3}[\/latex]<\/p>\n

9.\u00a0[latex]5[\/latex] units<\/p>\n

Solutions to Odd-Numbered Exercises<\/h2>\n

1.\u00a0It is a second-degree equation (the highest variable exponent is 2).<\/p>\n

3.\u00a0We want to take advantage of the zero property of multiplication in the fact that if [latex]a\\cdot b=0[\/latex] then it must follow that each factor separately offers a solution to the product being zero: [latex]a=0\\text{ }or\\text{ b}=0[\/latex].<\/p>\n

5.\u00a0One, when no linear term is present (no x<\/em> term), such as [latex]{x}^{2}=16[\/latex]. Two, when the equation is already in the form [latex]{\\left(ax+b\\right)}^{2}=d[\/latex].<\/p>\n

7.\u00a0[latex]x=6[\/latex], [latex]x=3[\/latex]<\/p>\n

9.\u00a0[latex]x=\\frac{-5}{2}[\/latex], [latex]x=\\frac{-1}{3}[\/latex]<\/p>\n

11.\u00a0[latex]x=5[\/latex], [latex]x=-5[\/latex]<\/p>\n

13.\u00a0[latex]x=\\frac{-3}{2}[\/latex], [latex]x=\\frac{3}{2}[\/latex]<\/p>\n

15.\u00a0[latex]x=-2[\/latex]<\/p>\n

17.\u00a0[latex]x=0[\/latex], [latex]x=\\frac{-3}{7}[\/latex]<\/p>\n

19.\u00a0[latex]x=-6[\/latex], [latex]x=6[\/latex]<\/p>\n

21.\u00a0[latex]x=6[\/latex], [latex]x=-4[\/latex]<\/p>\n

23.\u00a0[latex]x=1[\/latex], [latex]x=-2[\/latex]<\/p>\n

25.\u00a0[latex]x=-2[\/latex], [latex]x=11[\/latex]<\/p>\n

27.\u00a0[latex]x=3\\pm \\sqrt{22}[\/latex]<\/p>\n

29.\u00a0[latex]z=\\frac{2}{3}\\\\[\/latex], [latex]z=-\\frac{1}{2}[\/latex]<\/p>\n

31.\u00a0[latex]x=\\frac{3\\pm \\sqrt{17}}{4}[\/latex]<\/p>\n

33.\u00a0Not real<\/p>\n

35.\u00a0One rational<\/p>\n

37.\u00a0Two real; rational<\/p>\n

39.\u00a0[latex]x=\\frac{-1\\pm \\sqrt{17}}{2}[\/latex]<\/p>\n

41.\u00a0[latex]x=\\frac{5\\pm \\sqrt{13}}{6}[\/latex]<\/p>\n

43.\u00a0[latex]x=\\frac{-1\\pm \\sqrt{17}}{8}[\/latex]<\/p>\n

45.\u00a0[latex]x\\approx 0.131[\/latex] and [latex]x\\approx 2.535[\/latex]<\/p>\n

47.\u00a0[latex]x\\approx -6.7[\/latex] and [latex]x\\approx 1.7[\/latex]<\/p>\n

49.\u00a0[latex]\\begin{array}{l}a{x}^{2}+bx+c \\hfill& =0\\hfill \\\\ {x}^{2}+\\frac{b}{a}x \\hfill& =\\frac{-c}{a}\\hfill \\\\ {x}^{2}+\\frac{b}{a}x+\\frac{{b}^{2}}{4{a}^{2}} \\hfill& =\\frac{-c}{a}+\\frac{b}{4{a}^{2}}\\hfill \\\\ {\\left(x+\\frac{b}{2a}\\right)}^{2}\\hfill& =\\frac{{b}^{2}-4ac}{4{a}^{2}}\\hfill \\\\ x+\\frac{b}{2a}\\hfill& =\\pm \\sqrt{\\frac{{b}^{2}-4ac}{4{a}^{2}}}\\hfill \\\\ x\\hfill& =\\frac{-b\\pm \\sqrt{{b}^{2}-4ac}}{2a}\\hfill \\end{array}[\/latex]<\/p>\n

51.\u00a0[latex]x\\left(x+10\\right)=119[\/latex]; 7 ft. and 17 ft.<\/p>\n

53.\u00a0maximum at [latex]x=70[\/latex]<\/p>\n

55.\u00a0The quadratic equation would be [latex]\\left(100x - 0.5{x}^{2}\\right)-\\left(60x+300\\right)=300[\/latex]. The two values of [latex]x[\/latex] are 20 and 60.<\/p>\n

57.\u00a03 feet<\/p>\n\n\t\t\t

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Candela Citations<\/h3>\n\t\t\t\t\t
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