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##### Sections

###### Exercises

Exercise name | Free? |
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Exercises 1 We will start simplifying the given expression by splitting it into factors. Next, we will rewrite the number under the square root as the product of at least one perfect square factor. 112x3Split into factors16⋅7⋅x2⋅xCommutative Property of Multiplication16⋅x2⋅7⋅xa⋅b=a⋅b16⋅x2⋅7xCalculate root4⋅x⋅7⋅xMultiply4x7x | |

Exercises 2 In order to simplify the given expression, we will use the basic Properties of Radicals. Next, we will rewrite the number under the square root as the product of at least one perfect square factor. 8118ba=ba8118Split into factors819⋅2a⋅b=a⋅b819⋅2Calculate root93⋅2ba=b/3a/332 | |

Exercises 3 We can simplify this expression by rewriting the number under the cube root as the product of at least one perfect cube factor. Let's do it! 3-625Split into factors3-125⋅53a⋅b=3a⋅3b3-125⋅35Write as a power3(-5)3⋅35nan=a{n}-535 | |

Exercises 4 To simplify the given expression, let’s multiply the numerator and denominator by 32. This will eliminate the radical in the denominator. 3212ba=b⋅32a⋅3232⋅3212⋅32a⋅a=a3212⋅32 We know that we have successfully rationalized the denominator because the radical has been eliminated. However, our fraction can still be simplified a bit further. Let's try! 3212⋅32Split into factors3212⋅16⋅2a⋅b=a⋅b3212⋅16⋅2Calculate root3212⋅4⋅2Multiply32482ba=b/16a/16232 | |

Exercises 5 To rationalize a monomial denominator, we multiply the numerator and denominator by a radical that will eliminate the radical in the denominator. Because the radical in the denominator is a square root, we need to multiply it by a square root and that will give us a perfect square under the radical. 114ba=b⋅11a⋅1111⋅11411a⋅a=a11411 | |

Exercises 6 In order to simplify the given expression, we will use the basic Properties of Radicals. Next, we will rationalize the denominator. 13144ba=ba13144Calculate root1312ba=b⋅13a⋅1313⋅1312⋅13a⋅a=a1312⋅13Multiply131213 | |

Exercises 7 In order to simplify the given expression, we are going to use the Properties of Radicals. We will start by distributing the root and simplify the denominator. 3343y654x43ba=3b3a3343y6354x4Write as a power373y6354x4Split into factors373y2(3)354x4am⋅n=(am)n373(y2)3354x43a⋅b=3a⋅3b373⋅3(y2)3354x4nan=a{n}7y2354x4 Now that we have simplified the denominator, we are going to simplify the numerator. 7y2354x4Split into factors7y2327⋅2⋅x3⋅xCommutative Property of Multiplication7y2327x3⋅2xWrite as a power7y2333x3⋅2xambm=(ab)m7y23(3x)3⋅2x3a⋅b=3a⋅3b7y23(3x)332xnan=a{3}7y23x32x | |

Exercises 8 In order to simplify the given expression, we are first going to use the Quotient Property of Radicals. Let's do it! 28y4z54x2ba=ba28y4z54x2 To simplify the radicals, we should determine whether we need to use an absolute value symbol. Consider the following two cases for any real number a. nan={a if n is odd∣a∣ if n is even Because our radical has an even index, we will need to use an absolute value symbol for the variables that have even exponent to simplify our expression. 28y4z54x2Write as a sum28y4z1+44x2a1+m=a⋅am28y4zz44x2Split into factors4(7)y2(2)zz2(2)4x2Write as a power22(7)y2(2)zz2(2)22x2am⋅n=(am)n22(7)(y2)2z(z2)222x2Commutative Property of Multiplication22(y2)2(z2)2(7)z22x2ambm=(ab)m(2y2z2)2(7)z(2x)2a⋅b=a⋅b(2y2z2)27z(2x)2a2=∣a∣∣2y2z2∣7z∣2x∣ Note that, since 2, y2, and z2 are all nonnegative numbers, we can extract them from the absolute value symbol. ∣2y2z2∣7z∣2x∣⇔2y2z27z2∣x∣ Finally, we can simplify the algebraic fraction. 2y2z27z2∣x∣ba=b/2a/2y2z27z∣x∣ We can see that the denominator still involves a radical expression. In that case, let's rationalize the denominator. y2z27z∣x∣Multiply by 7z7zy2z27z∣x∣⋅7z7zMultiply fractionsy2z27z⋅7z∣x∣⋅7za⋅a=ay2z2⋅7z∣x∣⋅7zCommutative Property of Multiplication7y2zz2∣x∣7za⋅am=a1+m7y2z3∣x∣7z | |

Exercises 9 To rationalize a fraction with a binomial denominator, we multiply the numerator and denominator of the fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the second term of the expression.BinomialConjugate a+ba−b a−ba+bIn this case, the conjugate of the denominator is 5−3. 5+36ba=b⋅(5−3)a⋅(5−3)(5+3)(5−3)6(5−3)Distribute 6(5+3)(5−3)6⋅5−6⋅3Multiply(5+3)(5−3)30−63(a+b)(a−b)=a2−b252−(3)230−63(a)2=a25−330−63Subtract terms2230−63ba=b/2a/21115−33 | |

Exercises 10 To simplify the given radical expression, we first need to rewrite the radicands so that they have exponents that match the index of the radicals. Then we will consider the properties for combining radical expressions when they are part of a sum or difference. anx+bnx=(a+b)nxanx−bnx=(a−b)nx Notice that radicals can only be added or subtracted when the index and the value inside are exactly the same. Let's use perfect squares to simplify our radicals to see if we can create like terms. 25+710−320Split into factors25+75−34⋅5a⋅b=a⋅b25+710−34⋅5Calculate root25+710−3⋅2⋅5Multiply25+710−65Commutative Property of Addition25−65+710Factor out 5(2−6)5+710Subtract terms-45+710 | |

Exercises 11 To rationalize a fraction with a binomial denominator, we multiply the numerator and denominator of the fraction by the conjugate of the denominator. We find the conjugate by changing the sign of the second term of the expression.BinomialConjugate a+ba−b a−ba+bIn this case, the conjugate of the denominator is 8+10. 8−1010ba=b⋅(8+10)a⋅(8+10)(8−10)(8+10)10(8+10)Distribute 10(8−10)(8+10)108+1010(a+b)(a−b)=a2−b2((8)2−(10)2)108+1010(a)2=a8−10108+1010Subtract terms-2108+1010Factor out -2-2-2(-58−510)ba=b/(-2)a/(-2)1-58−5101a=a-58−510 Let's simplify the result further. -58−510Split into factors-52⋅2⋅2−510a⋅a=a2-522⋅2−510a⋅b=a⋅b-522⋅2−510a2=a-5⋅2⋅2−510Multiply-102−510 | |

Exercises 12 We will begin simplifying the expression by removing the parentheses. 6(712−43)Distribute 66⋅712−6⋅43Commutative Property of Multiplication7⋅6⋅12−4⋅6⋅3a⋅b=a⋅b76⋅12−46⋅3 Next, we need to rewrite the radicands so that they have exponents that match the index of the radicals. Then we will consider the properties for combining radical expressions when they are part of a sum or difference. anx+bnx=(a+b)nxanx−bnx=(a−b)nx Notice that radicals can only be added or subtracted when the index and the value inside are exactly the same. Let's use perfect squares to simplify our radicals to see if we can create like terms. 76⋅12−46⋅3Split into factors72⋅3⋅2⋅2⋅3−42⋅3⋅3Commutative Property of Multiplication72⋅2⋅3⋅3⋅2−43⋅3⋅2a⋅a=a2722⋅32⋅2−432⋅2a⋅b=a⋅b722⋅32⋅2−432⋅2a2=a7⋅2⋅3⋅2−4⋅3⋅2Multiply422−122Factor out 2(42−12)2Subtract terms302 | |

Exercises 13 We want to use the given graph to determine the solutions of the related quadratic equation. The solutions of this equation are the x-coordinates of the points of intersection of the parabola and the x-axis. Recall that a quadratic equation can have two, one, or no solutions. Let's consider the graph.This graph crosses the x-axis twice, so its related equation x2−2x−3 has two solutions. We can see that those solutions are x=-1 and x=3. Extra info Number of solutions shown on a graph window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21363_2_464057081_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var el = {}; var b1,b2,bb,xMax; b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); bb = b.board.getBoundingBox(); xMax = bb[2]; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph for two,}\\\\\\text{one, or no solutions!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); /*Reset*/ mlg.af("showD",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph for two,}\\\\\\text{one, or no solutions!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); }); /*A Two real solutions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#62d8ff", fontSize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#62d8ff", fontSize:0.8 } ); }); /*B One Real Solution: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#62ffd8", fontSize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#62ffd8", fontSize:0.8 } ); }); /*C No Real Solutions: orange #ff5116 and #ff8a62*/ mlg.af("showC",function() { b.remove(el); el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff5116'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff5116'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontSize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontSize:0.8 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21363_2_464057081_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21363_2_464057081_l", "Solution21363_2_464057081_p", 1, code); }); } ); } window.JXQtable["Solution21363_2_464057081_l"] = true;Two solutions window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn387683197_1910778555').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );One solution window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn138529284_948596759').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );No solutions window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1374109308_1952340771').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn801390792_2020973005').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); | |

Exercises 14 We want to use the given graph to determine the solutions of the related quadratic equation. The solutions of this equation are the x-coordinates of the points of intersection of the parabola and the x-axis. Recall that a quadratic equation can have two, one, or no solutions. Let's consider the graph.This graph does not cross the x-axis. Therefore, its related equation x2−2x+3 has no solutions. Extra info Number of solutions shown on a graph window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21364_21_117190561_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var el = {}; var b1,b2,bb,xMax; b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); bb = b.board.getBoundingBox(); xMax = bb[2]; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph for two,}\\\\\\text{one, or no solutions!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); /*Reset*/ mlg.af("showD",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph for two,}\\\\\\text{one, or no solutions!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); }); /*A Two real solutions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#62d8ff", fontSize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#62d8ff", fontSize:0.8 } ); }); /*B One Real Solution: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#62ffd8", fontSize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#62ffd8", fontSize:0.8 } ); }); /*C No Real Solutions: orange #ff5116 and #ff8a62*/ mlg.af("showC",function() { b.remove(el); el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff5116'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff5116'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontSize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontSize:0.8 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21364_21_117190561_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21364_21_117190561_l", "Solution21364_21_117190561_p", 1, code); }); } ); } window.JXQtable["Solution21364_21_117190561_l"] = true;Two solutions window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn201811492_1250465866').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );One solution window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn2116060529_1655021538').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );No solutions window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1791904722_2069080249').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1606331435_1907868809').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); | |

Exercises 15 We want to use the given graph to determine the solutions of the related quadratic equation. The solutions of this equation are the x-coordinates of the points of intersection of the parabola and the x-axis. Recall that a quadratic equation can have two, one, or no solutions. Let's consider the graph.This graph intercepts the x-axis at one point, so its related equation x2+10x+25 has one solution. We can see that the solution is x=-5. Extra info Number of solutions shown on a graph window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution2136_2_691892659_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { //board elements var b = mlg.board([-11,11,11,-11],{desktopSize:'large',"style":"usa"}); var el = {}; var b1,b2,bb,xMax; b1 = b.subBoard([0,0],[-1.75,1.75,1.75,-1.75]); b2 = b.subBoard(b1,[-1.75,1.75,1.75,-1.75],{padding:0.1}); b.subCrop([b1,b2],0); bb = b.board.getBoundingBox(); xMax = bb[2]; //default display b1.xaxis(2,0); b1.yaxis(2,0); b2.xaxis(2,0); b2.yaxis(2,0); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph for two,}\\\\\\text{one, or no solutions!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); /*Reset*/ mlg.af("showD",function() { b.remove(el); el.txt0 = b.txt(0.33*xMax,0, "\\begin{gathered}\\text{See the graph for two,}\\\\\\text{one, or no solutions!}\\end{gathered}", { mathMode:true, flag:true, fontSize:1.2, layer:100 } ); }); /*A Two real solutions: blue #628aff and #62d8ff*/ mlg.af("showA",function() { b.remove(el); el.p1 = b1.point(1.22,0,{size:0.25}); el.p2 = b1.point(-1.22,0,{size:0.25}); el.p3 = b2.point(1.22,0,{size:0.25}); el.p4 = b2.point(-1.22,0,{size:0.25}); el.f1 = b1.func('x^2-1.5',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.f2 = b2.func('-1*x^2+1.5',{strokeWidth:1.8,strokeColor:'#628aff',xMax:1.7,xMin:-1.7}); el.txt1 = b1.txt(0,0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#62d8ff", fontSize:0.8 } ); el.txt2 = b2.txt(0,-0.9,"Two Real Solutions",{ mathMode:false, flag:true, flagColor:"#62d8ff", fontSize:0.8 } ); }); /*B One Real Solution: green #00c997 and #62ffd8*/ mlg.af("showB",function() { b.remove(el); el.p1 = b1.point(0,0,{size:0.25}); el.p2 = b2.point(0,0,{size:0.25}); el.f1 = b1.func('x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.f2 = b2.func('-1*x^2',{strokeWidth:1.8,strokeColor:'#00c997'}); el.txt1 = b1.txt(0,-0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#62ffd8", fontSize:0.8 } ); el.txt2 = b2.txt(0,0.9,"One Real Solution",{ mathMode:false, flag:true, flagColor:"#62ffd8", fontSize:0.8 } ); }); /*C No Real Solutions: orange #ff5116 and #ff8a62*/ mlg.af("showC",function() { b.remove(el); el.f1 = b1.func('x^2+0.5',{strokeWidth:1.8,strokeColor:'#ff5116'}); el.f2 = b2.func('-1*x^2-0.5',{strokeWidth:1.8,strokeColor:'#ff5116'}); el.txt1 = b1.txt(0,-0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontSize:0.8 } ); el.txt2 = b2.txt(0,0.9,"No Real Solutions",{ mathMode:false, flag:true, flagColor:"#ff8a62", fontSize:0.8 } ); }); } catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution2136_2_691892659_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution2136_2_691892659_l", "Solution2136_2_691892659_p", 1, code); }); } ); } window.JXQtable["Solution2136_2_691892659_l"] = true;Two solutions window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1970298739_880761002').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );One solution window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn2103752404_942412874').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );No solutions window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn37038907_2124858028').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );Reset window.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn260740465_1041660904').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } ); | |

Exercises 16 To solve the equation by graphing, we will consider the left-hand side as a function. To draw the graph of a quadratic function written in standard form, we must start by identifying the values of a, b, and c. y=x2+9x+14⇕y=1x2+9x+14 We can see that a=1, b=9, and c=14. Now, we will follow four steps to graph the function.Find the axis of symmetry. Calculate the vertex. Identify the y-intercept and its reflection across the axis of symmetry. Connect the points with a parabola.Finding the Axis of Symmetry The axis of symmetry is a vertical line with equation x=-2ab. Since we already know the values of a and b, we can substitute them into the formula. x=-2aba=1, b=9x=-2(1)9Multiplyx=-29Calculate quotientx=-4.5 The axis of symmetry of the parabola is the vertical line with equation x=-4.5.Calculating the Vertex To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: (-2ab,f(-2ab)) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=-4.5. Thus, the x-coordinate of the vertex is also -4.5. To find the y-coordinate, we need to substitute -4.5 for x in the given equation. y=x2+9x+14x=-4.5y=(-4.5)2+9(-4.5)+14 Simplify right-hand side Calculate powery=20.25+9(-4.5)+14Multiplyy=20.25−40.5+14Add and subtract terms y=-6.25 We found the y-coordinate, and now we know that the vertex is (-4.5,-6.25).Identifying the y-intercept and Its Reflection The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,14). Let's plot this point and its reflection across the axis of symmetry.Connecting the Points We can now draw the graph of the function. Since a=1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.From the graph we can see that y=x2+9x+14 is equal to 0 for x=-7 and x=-2. That gives us that x=-7 and x=-2 are the solutions to the given equation. Checking Our Answer info Checking Our Solutions We can check our solutions by substituting the values into the given function. If our solutions are correct, the final result will be 0=0. Let's first check x=-7. x2+9x+14=0x=-7(-7)2+9(-7)+14=?0Calculate power49+9(-7)+14=?0Multiply49−63+14=?0Add and subtract terms0=0✓ Great! Now, let's check our second value, x=-2. x2+9x+14=0x=-2(-2)2+9(-2)+14=?0Calculate power4+9(-2)+14=?0Multiply4−18+14=?0Add and subtract terms0=0✓ | |

Exercises 17 In order to solve the equation, we will rewrite it so that all the terms are on the left-hand side. x2−7x=8⇔x2−7x−8=0 Now we will consider the left-hand side as a quadratic function written in standard form. Let's identify the values of a, b, and c. y=x2−7x−8⇕y=1x2+(-7)x+(-8) We can see that a=1, b=-7, and c=-8. Now, we will follow four steps to graph the function.Find the axis of symmetry. Calculate the vertex. Identify the y-intercept and its reflection across the axis of symmetry. Connect the points with a parabola.Finding the Axis of Symmetry The axis of symmetry is a vertical line with equation x=-2ab. Since we already know the values of a and b, we can substitute them into the formula. x=-2aba=1, b=-7x=-2(1)-7 Simplify right-hand side Identity Property of Multiplicationx=-2-7-b-a=bax=27Calculate quotient x=3.5 The axis of symmetry of the parabola is the vertical line with equation x=3.5.Calculating the Vertex To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: (-2ab,f(-2ab)) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=3.5. Thus, the x-coordinate of the vertex is also 3.5. To find the y-coordinate, we need to substitute 3.5 for x in our function. y=x2−7x−8x=3.5y=(3.5)2−7(3.5)−8 Simplify right-hand side Calculate powery=12.25−7(3.5)−8Multiplyy=12.25−24.5−8Subtract terms y=-20.25 We found the y-coordinate, and now we know that the vertex is (3.5,-20.25).Identifying the y-intercept and Its Reflection The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,-8). Let's plot this point and its reflection across the axis of symmetry.Connecting the Points We can now draw the graph of the function. Since a=1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.By looking at the graph we can see that the x- intercepts, thus solutions to the equation, are x=-1 and x=8. Checking Our Answer info Checking Our Solutions We can check our solutions by substituting the values into the given function. If our solutions are correct, the final result will be 0=0. Let's first check x=-1. x2−7x−8=0x=-1(-1)2−7(-1)−8=?0Calculate power1−7(-1)−8=?0Multiply1+7−8=?0Add and subtract terms0=0 ✓ We have confirmed that -1 is a solution to the equation. Now, let's check our second value, x=8. x2−7x−8=0x=8(8)2−7(8)−8=?0Calculate power64−7(8)−8=?0Multiply64−56−8=?0Add and subtract terms0=0 ✓ Now we know that 8 is also a solution to the equation. | |

Exercises 18 In order to solve the equation, we will rewrite it so that all the terms are on the left-hand side. x+4=-x2⇔x2+x+4=0 Now we will consider the left-hand side as a quadratic function written in standard form. Let's identify the values of a, b, and c. y=x2+x+4=0⇕y=1x2+1x+4 We can see that a=1, b=1, and c=4. Now, we will follow four steps to graph the function.Find the axis of symmetry. Calculate the vertex. Identify the y-intercept and its reflection across the axis of symmetry. Connect the points with a parabola.Finding the Axis of Symmetry The axis of symmetry is a vertical line with equation x=-2ab. Since we already know the values of a and b, we can substitute them into the formula. x=-2aba=1, b=1x=-2(1)1a⋅1=ax=-21Calculate quotientx=-0.5 The axis of symmetry of the parabola is the vertical line with equation x=-0.5.Calculating the Vertex To calculate the vertex, we need to think of y as a function of x, y=f(x). We can write the expression for the vertex by stating the x- and y-coordinates in terms of a and b. Vertex: (-2ab,f(-2ab)) Note that the formula for the x-coordinate is the same as the formula for the axis of symmetry, which is x=-0.5. Thus, the x-coordinate of the vertex is also -0.5. To find the y-coordinate, we need to substitute -0.5 for x in the given equation. y=x2+x+4x=-0.5y=(-0.5)2+(-0.5)+4Calculate powery=0.25+(-0.5)+4Add and subtract termsy=3.75 We found the y-coordinate, and now we know that the vertex is (-0.5,3.75).Identifying the y-intercept and Its Reflection The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,4). Let's plot this point and its reflection across the axis of symmetry.Connecting the Points We can now draw the graph of the function. Since a=1, which is positive, the parabola will open upwards. Let's connect the three points with a smooth curve.By looking at the graph, we can state that there are no x-intercepts, therefore there are no real solutions to given equation. | |

Exercises 19 To solve the given equation by taking the square roots, we need to consider the positive and negative solutions. 4x2=64LHS/4=RHS/4x2=16LHS=RHSx=±16Calculate rootx=±4 We found that x=±4. Thus, there are two solutions for the equation x=4 and x=-4. Checking Our Answer info Checking our answer We can check our answers by substituting them for x in the given equation. Let's start with x=-4. 4x2=64x=-44(-4)2=?64 Simplify (-a)2=a24(4)2=?64Calculate power4(16)=?64Multiply 64=64 ✓ We have confirmed that -4 is a correct solution. Let's now check if x=4. 4x2=64x=44(4)2=?64Calculate power4(16)=?64Multiply64=64 ✓ Now we know that x=4 is also a solution of the equation. | |

Exercises 20 Let's begin by isolating x2 from the equation and then consider if there exists any real solutions. -3x2+6=10LHS−6=RHS−6-3x2=4LHS/(-3)=RHS/(-3)x2=-34 The square of a real number cannot be negative. Therefore, the equation has no real solutions. | |

Exercises 21 To solve the given equation by taking the square roots, we need to consider the positive and negative solutions. (x−8)2=1LHS=RHSx−8=±1Calculate rootx−8=±1State solutionsx−8=1x−8=-1(I)(II)(I), (II): LHS+8=RHS+8x=9x=7 We found that there are two solutions for the equation, x=9 and x=7. Checking Our Answer info Checking our answer We can check our answers by substituting them for x in the given equation. Let's start with x=9. (x−8)2=1x=9(9−8)2=?1Subtract terms12=?1Calculate power1=1✓ Now we know that x=9 is a solution of the equation. Let's check if x=7 is also a solution. (x−8)2=1x=7(7−8)2=?1Subtract terms(-1)2=?1Calculate power1=1✓ This gives us that x=7 is also a solution of the equation. | |

Exercises 22 Let's start by reviewing how can we find the solutions to an equation of the form x2=d by using square roots. For this, we can take the square root of each side of the equation to isolate x. Notice that there are three possible cases according to the value of d.Case d>0. In this case, since d=-d, the equation x2=d has two real solutions, x=±d. Case d=0. In this case, since 0=-0, the equation x2=d has one real solution, x=0. Case d<0. In this case, since d is not a real number, the equation x2=d has no real solutions.We want to determine the number of real solutions for the equation shown below. x2=100 Therefore, since 100>0, we can know that the equation has 2 real solutions. | |

Exercises 23 Let's recall the formula for the volume of a rectangular prism. V=ℓwh Here ℓ is the length of the prism, w is the width of the prism, and h is its height. In our case V=380 and h=5. We also know that the length ℓ is four times the width w, so ℓ=4w. We will substitute these values into the formula. By doing so we obtain an equation that can be solved for w. V=ℓwhSubstitute V=380,h=5,ℓ=4w380=4w(w)(5)Multiply380=20w2LHS/20=RHS/2019=w2LHS=RHS±19=wRearrange equationw=±19 The solutions to the equation are 19 and -19. Since w represents the width of the prism, it cannot be negative. Therefore, w=19 and ℓ=419. We can use a calculator to approximate our results. wℓ=19≈4.4=419≈17.4 The width is 19≈4.4 meters and the length is 419≈17.4 meters. | |

Exercises 24 |

##### Other subchapters in Solving Quadratic Equations

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Properties of Radicals
- Solving Quadratic Equations by Graphing
- Solving Quadratic Equations Using Square Roots
- Solving Quadratic Equations by Completing the Square
- Solving Quadratic Equations Using the Quadratic Formula
- Solving Nonlinear Systems of Equations
- Chapter Review
- Chapter Test
- Cumulative Assessment