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Exercises 1 The information whether the discriminant is positive, negative, or zero can be used to determine the number of real solutions of a quadratic equation. However, it works both ways — the number of real solutions tells us whether the discriminant is positive, negative, or zero.DiscriminantNumber of Solutions PositiveTwo real solutions NegativeNo real solutions ZeroOne real solutionTo determine whether the discriminants of the given equations are positive, negative, or zero, we will determine the number of solutions of each equation. f(x)=0g(x)=0h(x)=0j(x)=0 The solutions of these equations, if any, are the x-intercepts of the related functions, f, g, h, and j, respectively. Let's take a look at the given graph.window.JXQ = window.JXQ || []; window.JXQtable = window.JXQtable || {}; if(!window.JXQtable["Solution21713_4335870_l"]) { window.JXQ.push( function () { var code = function (gid) { var images = []; try { var b=mlg.board([-7.5,6.5,7.5,-6.5],{desktopSize:"medium",style:"usa"}); b.xaxis(2,1); b.yaxis(2,1); var el = {};el.func1 = b.func("1.4*x^2+6.1*x+7.7"); el.func2 = b.func("-2*x^2-3*x+2",{strokeColor:"red"}); el.func3 = b.func("x^2-4*x+4",{strokeColor:"green"}); el.func4 = b.func("-0.32*x^2",{strokeColor:"darkviolet"}); el.legend1 = b.legend(el.func1,[-3.91,5.03],"f"); el.legend2 = b.legend(el.func2,[-2.63,-3.93],"g"); el.legend3 = b.legend(el.func3,[3.54,2.07],"h"); el.legend4 = b.legend(el.func4,[3.14,-2.89],"j");mlg.af("showA",function() { b.remove(el); el.func1 = b.func("1.4*x^2+6.1*x+7.7"); el.func2 = b.func("-2*x^2-3*x+2",{strokeColor:"red"}); el.func3 = b.func("x^2-4*x+4",{strokeColor:"green"}); el.func4 = b.func("-0.32*x^2",{strokeColor:"darkviolet"}); el.legend1 = b.legend(el.func1,[-3.91,5.03],"f"); el.legend2 = b.legend(el.func2,[-2.63,-3.93],"g"); el.legend3 = b.legend(el.func3,[3.54,2.07],"h"); el.legend4 = b.legend(el.func4,[3.14,-2.89],"j"); });mlg.af("showB",function() { b.remove(el); el.func1 = b.func("1.4*x^2+6.1*x+7.7"); el.legend1 = b.legend(el.func1,[-3.91,5.03],"f"); });mlg.af("showC",function() { b.remove(el); el.func2 = b.func("-2*x^2-3*x+2",{strokeColor:"red"}); el.legend2 = b.legend(el.func2,[-2.63,-3.93],"g"); el.p1 = b.point(-2,0); el.p2 = b.point(0.5,0); el.flag1 = b.flag(el.p1,"x\\text{-intercept}",152,2.3); el.flag2 = b.flag(el.p2,"x\\text{-intercept}",27,2.3); });mlg.af("showD",function() { b.remove(el); el.func3 = b.func("x^2-4*x+4",{strokeColor:"green"}); el.func4 = b.func("-0.32*x^2",{strokeColor:"darkviolet"}); el.legend3 = b.legend(el.func3,[3.54,2.07],"h"); el.legend4 = b.legend(el.func4,[3.14,-2.89],"j"); el.p1 = b.point(2,0,{fillColor:"green"}); el.p2 = b.point(0,0,{fillColor:"darkviolet"}); el.flag1 = b.flag(el.p1,"x\\text{-intercept}",19,3.9); el.flag2 = b.flag(el.p2,"x\\text{-intercept}",161,3.4); });} catch( e ) { mw.hook("ml.error.report").fire(e, { tags: { module: "jsxg", graph: "Solution21713_4335870_l" } }); console.error(e); } if(b) {b.board.update()} }; mw.loader.using('ext.MLJSXGraph', function() { window.ml.jsxgraph.buildRealGraph("Solution21713_4335870_l", "Solution21713_4335870_p", 1, code); }); } ); } window.JXQtable["Solution21713_4335870_l"] = true;Allwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1192738778_904360145').on('touchstart mousedown', function () { try { mlg.cf("showA"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );fwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1207289197_1730546574').on('touchstart mousedown', function () { try { mlg.cf("showB"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );gwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn1113131442_1865803475').on('touchstart mousedown', function () { try { mlg.cf("showC"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );h and jwindow.RLQ = window.RLQ || []; window.RLQ.push( function () { (function ($) { $('#jsxbtn644397787_566050357').on('touchstart mousedown', function () { try { mlg.cf("showD"); } catch(e) { mw.log.error(e); } }); })(jQuery); } );We can see that f has no x-intercepts, g has two x-intercepts, and h and j have one x-intercept each. Therefore, f(x)=0 has no real solutions, g(x)=0 has two real solutions, and h(x)=0 and j(x)=0 have one real solution each. Let's make a table to see what it means in terms of the dicriminants!EquationNumber of SolutionsDiscriminant f(x)=0No real solutionsNegative g(x)=0Two real solutionsPositive h(x)=0One real solutionZero j(x)=0One real solutionZero | |

Exercises 2 | |

Exercises 3 | |

Exercises 4 We are given two data sets. We will graph each of them to determine whether each table shows a positive, a negative, or no correlation.Graph DescriptionCorrelation y increases as x increasesPositive y decreases as x increasesNegative There is no relationship between x and yNone First, let's consider the table showing the number of cups of hot chocolate sold.Temperature (oF), xCups of hot chocolate, y 1435 2728 3222 419 484 622 731 We will plot the coordinate pairs (x,y). Remember to choose an appropriate scale for the x-axis and the y-axis, so that all the points fit on the graph.We can see that as the value of x increases the value of y decreases. The points follow a line that is trending down. Therefore, the data represented by the table shows negative correlation. Next, let's consider the table showing the number of bottles of sports drink sold.Temperature (oF), xBottles of sports drink, y 148 2712 3213 4116 4819 6227 7329 Once more we will plot the coordinate pairs (x,y).We can see that as the value of x increases the value of y also increases. The points follow a line that is trending up. Therefore, the data represented by the table shows positive correlation. | |

Exercises 5 An exponential growth function can be written in the following form. y=a⋅bx,a>0 and b>1 The point (0,a) is the y-intercept and the start of the graph. Since the base of the power b is greater than 1, the quantity grows faster and faster without bound. Thus, the graph of an exponential growth function gently curves up and approaches a vertical line. Now let's consider the given graphs.GraphDescriptionExponential Growth? AGently curves up and approaches a vertical lineYes ✓ BGently curves down and approaches a horizontal lineNo × CStraight lineNo × DU-shapedNo × We can see that only the description of graph A matches this of a graph of an exponential growth. Therefore, the answer is A. | |

Exercises 6 Let's take a look at the graph representing the given system of equations.We can see that the graphs intersect at two points, (-2,-8) and (5,27). Therefore, the system has two solutions. Our answer corresponds to option B. Showing Our Work info Graphing In case you want to see the steps we took to graph the system, we have included the process. First we graphed the function y=x2+2x−8. We began by determining the values of a, b, and c. y=x2+2x−8⇔y=1x2+2x+(-8) Therefore, a=1, b=2, and c=-8. Then we identified the axis of symmetry. x=-2aba=1, b=2x=-2(1)2 Simplify ba=b/2a/2x=-11aa=1 x=-1 The axis of symmetry for this function is x=-1. Next we determined the vertex. The axis of symmetry always intersects the parabola at its vertex. Therefore, x=-1 is the x-coordinate of the vertex. We used this to determine the corresponding y-coordinate. y=x2+2x−8x=-1y=(-1)2+2(-1)−8 Simplify (-a)2=a2y=1+2(-1)−8a(-b)=-a⋅by=1−2−8Subtract terms y=-9 The vertex of the function is (-1,-9). We then found one more point that lies on the parabola. To do so we used x=4. y=x2+2x−8x=4y=42+2(4)−8 Simplify Calculate powery=16+2(4)−8Multiplyy=16+8−8Add and subtract terms y=16 The point (4,16) lies on the parabola. Finally, we plotted the axis of symmetry x=-1, the vertex (-1,-9), and the point (4,16). We also reflected this point across the axis of symmetry.Finally we connected the three points with a smooth curve.Next we graphed the linear function y=5x+2. We identified the values of the slope m and the y-intercept b. y=5x+2 We plotted the y-intercept and used the slope to plot another point. Then we connected the points with a straight edge.The last step was connecting the graphs. | |

Exercises 7 We are asked to determine which expressions are in the simplest form.We will analyze the expressions one at a time to check if they can be further simplified.Expression x45x Let's take a look at the expression x45x. The expression under the radical sign 45x can be factored. Furthermore, it can be factored so that one of the factors is a perfect square. 45x=9⋅5x The factor 9 is a perfect square. Let's use this information to simplify our expression. x45xRewrite 45x as 9⋅5xx9⋅5xa⋅b=a⋅bx⋅9⋅5xCalculate powerx⋅3⋅5xMultiply3x5x Therefore, the given expression is not written in the simplest form.Expression 516 The denominator of the expression 516 can be rationalized. 516ba=b⋅5a⋅55⋅516⋅5a⋅a=a516⋅5Multiply5165 Therefore, the given expression is not written in the simplest form.Expression 394 The expression 394 cannot be simplified.The fraction cannot be reduced. The expression under the radical sign is not a perfect cube, nor does it have factors that are perfect cubes.Therefore, the given expression is written in the simplest form.Expression 165 The expression 165 cannot be simplified, because 5 is not a perfect square nor does it have factors that are perfect squares. Therefore, the given expression is written in the simplest form.Expression 3x5x The expression 3x5x cannot be simplified, because 5x is not a perfect square nor does it have factors that are perfect squares. Therefore, the given expression is written in the simplest form.Expression 23x4 The numerator of the expression 23x4 can be simplified. 23x4Rewrite x4 as x3⋅x23x3⋅x3a⋅b=3a⋅3b23x3⋅3xnan=a{n}2x⋅3xMultiply2x3x Therefore, the given expression is not written in the simplest form.Expression 347 The expression 347 cannot be simplified.The denominator is rational. The fraction cannot be reduced. The expression under the radical sign cannot be factored. The expression under the radical sign is not a perfect square.Therefore, the given expression is written in the simplest form.Expression 516 The numerator of the expression 516 is a perfect square. 16=42 This means we can simplify our expression. 516Rewrite 16 as 42542a2=a54 Therefore, the given expression is not written in the simplest form.Expression 23x2 The expression 23x2 cannot be simplified, because x2 is not a perfect cube nor does it have factors that are perfect cubes. Therefore, the given expression is written in the simplest form.Expression 3x7 The denominator of the expression 3x7 can be rationalized. 3x7ba=b⋅xa⋅x3x⋅x7⋅xa⋅a=a3x7⋅xa⋅b=a⋅b3x7x Therefore, the given expression is not written in the simplest form. | |

Exercises 8 We are asked to find all the ordered pairs (x,y) that are the solutions of the equation y=f(x), where f(x)=4x−5. y=f(x)⇔y=4x−5 Note that the domain of the function f(x) is all integers in the interval -3<x≤3. In other words, the domain is all integers that are greater than -3 and less than or equal to 3. Let's list them. -2,-1,0,1,2,3 By substituting these values of x into the equation y=4x−5 we will get the corresponding values of y. Then we will be able to determine the ordered pairs (x,y) that are the solutions of the equation y=f(x).x4x−5y -24(-2)−5-13 -14(-1)−5-9 04(0)−5-5 14(1)−5-1 24(2)−53 34(3)−57 Finally we will list the ordered pairs (x,y) formed in the table. (-2,-13),(-1,-9),(0,-5),(1,-1),(2,3),(3,7) These pairs are the solutions of the given equation y=f(x). |

##### 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
- Quiz
- Solving Quadratic Equations by Completing the Square
- Solving Quadratic Equations Using the Quadratic Formula
- Solving Nonlinear Systems of Equations
- Chapter Review
- Chapter Test