Principles of Mathematics Section, Introduction 03 Introduction Module, Section Solving Equations In this section, you will learn to solve quadratic equations graphically, by factoring, and by applying the quadratic formula. You will learn to draw connections between the zeros of a quadratic function, the -intercepts of the graph of the quadratic function, and the roots of the quadratic equation. Furthermore, you will become familiar with the nature of the roots of a quadratic equation by using the discriminant of a quadratic equation. Other types of equations that you will learn to solve are radical equations, rational equations, and absolute value equations. Section Outline Lesson Lesson Lesson 3 Lesson Lesson 5 Lesson 6 Solving Quadratic Equations Graphically Solving Quadratic Equations by Factoring Solving Quadratic Equations by Completing the Square and by the Quadratic Formula The Nature of Roots Radical Equations Rational and Absolute Value Equations Review Module
0 Section, Introduction Principles of Mathematics Notes Module
Principles of Mathematics Section, Lesson 05 Lesson Solving Quadratic Equations Graphically Outcomes When you complete this lesson, you will be able to: solve quadratic equations graphically and draw connections between the zeros of a quadratic function, the -intercepts of the graph of a quadratic function, and the roots of the quadratic equation solve quadratic equations of the form a + b + c = 0 solve quadratic equations where the solution is found in the set of imaginary numbers The Quadratic Equation Definition A Quadratic Equation is one which can be epressed in the form a + b + c = 0 where a 0 and a, b, and c are Real numbers. a + b + c = 0 is called the general form of the equation. The roots or solutions for a quadratic equation are simply the values of the variable which make the equation a true mathematical statement. For eample, is a root of the quadratic equation + 5 6 = 0 because () + 5() 6 = + 5 6 = 0. A given Quadratic Equation might have as many as two roots (solutions) or it may have no roots (solutions) at all! The Corresponding Quadratic Function and Its Graph For any Quadratic Equation a + b + c = 0, there is a corresponding Quadratic Function (which we analyzed in detail in Module, Section ) f() = a + b + c. The graph of this function was a parabola. The roots of the quadratic equation are the zeros of the corresponding quadratic function. These, in turn, correspond to the -intercepts of the parabola which is its graph. In this lesson we will find the roots (solutions) of some quadratic equations by graphing its corresponding function and inspecting its -intercepts. Module
06 Section, Lesson Principles of Mathematics Eample Determine the roots of the equation 3 =. Solution Our first step is to put the equation into general form, a + b + c = 0. 3 = 0. In this case a = 3, b = 0, and c = -. In order to graph the corresponding function y = 3 we first need the verte. In Module, Section we studied two methods of determining the verte. We can either complete the square, placing the function in standard form y = a( h) + k and read the verte (h,k) directly from the equation, or we can use the formula for the verte developed in Lesson 7. In these eamples we will use the latter method. Because the function y = 3 represents a vertical stretch (by a factor of 3) of the standard parabola y =, the formation is as follows: y ()( ) () b ac b 0 3 0 V, = V, = () a a 3 3 Now we sketch the graph. Five points are usually sufficient for the sketch of a parabola. 3 3 } } Module
Principles of Mathematics Section, Lesson 07 y 3 5 By inspection of the curve, we determine that the - intercepts are - and. Therefore the roots (solutions) for the original equation are - and. It's always a good idea to check these values in the original equation...just to be sure. 0 V(0, ) LHS RHS 3 3(-) 3() LHS RHS 3 3() 3() Since both of the solutions check, we can state that the Solution Set = { -, }. Module
08 Section, Lesson Principles of Mathematics Eample Solve the equation = -. Solution Our first step is to put the equation into general form, a + b + c = 0 re-arrange it to + = 0. In this case a =, b =, and c = 0. In order to graph the corresponding function y = + we need the verte. ()() () b ac b 0 V, = V, = -,- () a a ( ) The graph is standard in size (because the leading coefficient is ), so we start with formation. 5 3 3 0 y 3 Then using the verte (, ), the graph is shown below: y 5 3 3 3 The -intercepts of the corresponding quadratic function are -, and 0. Therefore the Solution Set = { -, 0 }. As an eercise you can check these values in the original equation. Module
Principles of Mathematics Section, Lesson 09 Q: Why does this equation use negative signs with two different lengths? A: On the T-83 there are two keys for - a short one in parenthesis (-) to indicate negative as in - or - and a longer one to indicate subtraction as in 5 3 or 6. The short one is only used before anumber or variable and never to operate on two terms. The longer one must be used between two terms. When writing equations by hand, you may ignore the difference completely! When using a graphing calculator, you have to follow that rule or the calculator will report an error. Eample 3 - Solve the equation =. Solution First rearrange the original equation so that it's in general form. - = - - + = 0 Now define y = - +. In this case a = -, b =, and c = -. Determine the Verte: The graph of this function is obtained by flipping (reflecting) the standard y = over (through) the ais, and then stretching it vertically by a factor of. So the formation will be as shown below. y } } 8 8 ( )( ) ( ) b ac b V, = V, =,0 ( ) a a ( ) The graph is shown on the net page: Module
0 Section, Lesson Principles of Mathematics y 3 3 There is only one - intercept in this case, so the quadratic equation has only one root (solution), namely. Solution Set is singleton set { }. 8 Of course, not all Quadratic Equations have integer solutions. If one of its solutions is nonintegral (in other words, it's not an integer) then manually graphing the corresponding function will not give you a very accurate answer. The Graphing Calculator, however, can give you a very close approimation. If you have a TI-83 (or the Langara Graph Eplorer) you can follow along with our final eample. Eample Solve the equation ( + ) = 5. Solution: ( + ) = 5 + = 5 + 5 = 0 Define Y = + 5 Press Y=. Then type X,T, θ,n + X,T, θ,n 5 Press GRAPH. You can see that the -intercepts ar e not integers. Press nd TRACE to access the "CALCULATE" menu. Press to locate your zeros (roots, -intercep ts). Module
Principles of Mathematics Section, Lesson Other graphing calculators have similar methods of determining the value of a noninteger root. Using the Langara Graph Eplorer, you have to zoom in to the - intercept to find it s approimate value. You will see your graph and be prompted for a Left Bound. Using the left arrow key locate the cursor just to the left of the point of intersection of the parabola and the -ais. ( = -.5539 is a good place). Press ENTER to confirm. Now you will see a prompt for a Right Bound. This time use the right arrow key to locate the cursor just to the right of the same point of intersection. ( = -.708 is a good place). Press ENTER to confirm. You are now prompted for a Guess. Again using the left arrow key, locate the cursor as close as possible to being on the actual point of intersection and press ENTER to confirm. Voilà! At the bottom of your screen your zero will appear, = -.583 which we will round to -.6. In a similar manner you will determine that the second zero is.583 which we will round to.6. So the Solution Set for the given equation is { -.6,.6 }. Now it's your turn to try a few. In the assignment which follows, # g) will require a graphing calculator solution. If you have one you can use it for that solution and you can also use it to check your other hand-drawn graphs. If you do not have one you may omit # g). Self-Marking Activity. Rearrange each of the Quadratic Equations so that they are in general form. a) b) c) d) e) f) g) + 0 = 6 3+ = 8 b 5 7+ 3 = 7 b b g g + 3 = g + 3 = 5 = 3 3 + = 0 b gb g Module
Section, Lesson Principles of Mathematics. Solve each of the following by manually graphing the corresponding quadratic function. You should use your graphing calculator for (g), if you have one. Use the blank grids that follow. a) + 8 = 0 b) c) + + 3 = 0 + 8 = 5 d) 9 = 0 e) + 0 = 0 f) -0 = + ( ) g)-= 3 a) y 8 6 6 8 Module
Principles of Mathematics Section, Lesson 3 b) y 8 6 6 8 c) y 8 6 6 8 Module
Section, Lesson Principles of Mathematics d) y 8 6 6 8 e) y 8 6 6 8 Module
Principles of Mathematics Section, Lesson 5 f) y 8 6 6 8 g) y 8 6 6 8 Check your answers in the Module Answer Key. Module
6 Section, Lesson Principles of Mathematics Module
Principles of Mathematics Section, Lesson 7 Lesson Solving Quadratic Equations by Factoring Outcome When you complete this lesson, you will be able to solve quadratic equations by factoring and by drawing connections among the zeros of functions, -intercepts of the graph, and roots of the quadratic equation. Overview You have solved quadratic equations by graphing. In some cases only approimate roots can be determined. However, it is possible to find the roots of some quadratics by factoring. This solution method is based on the principle of zero products from Module, Section. When using this property to solve quadratic equations, make sure the equation is in general form (with one side equal to zero). In Lesson you eamined a quadratic equation of the form a + b + c = 0, where b = 0. You solved this equation graphically. In this lesson we will solve the same type of equation by factoring. Recall from Principles of Mathematics 0 (and the Review in Module, Section ) that the difference of squares A B factors as (A B)(A + B). Eample Solve 9 = 0 by factoring. Solution ( )( ) 3 + 3 = 0 (A difference of two squares.) If the product of two epressions is equal to zero, then one of the two epressions must be zero. 3 = 0 or + 3 = 0 = 3 or = 3 3 3 = = You will use factoring to solve some equations of the type a + b = 0 or a + b + c = 0. Module
8 Section, Lesson Principles of Mathematics Eample Solve by factoring: a) 3 = 7 Solution 3 7 = 0 b 3 7 = 0 = 0 = 0 = 0 Check and verify these roots. b) 6 = 0 g Solution or or or ( 8) = 0 = 0 = 0 or or Set up equation in general form Factor out the common factor of 3 7 = 0 3 = 7 = 7 3 Factor out common factor of 8= 0 = 8 Check and verify roots. Eample 3 Solve: a) 3 + = Solution 3 + = 0 Set up in general form b3 gb + g = 0 Factor [Recall Trinomial Factoring from Principles of Math 0 or 3 = 0 = 3 or or + = 0 = Module, Section.] Check these roots. Module
Principles of Mathematics Section, Lesson 9 b) + 9 = 6 Solution 6 + 9 = 0 b gb g 3 3 = 0 3= 0 or = 3 or 3= 0 = 3 The answer is 3. When the same number repeats as a root or a zero, it is called a double root or double zero. If you graph 6 + 9 = 0, notice that the verte is on the -ais and the root is the same number as the -coordinate of the verte. c) + 6 = 0 Solution d i b gb g b + 8 = 0 + = 0 Either Check roots. d) 3 + 5 6 = 0 Solution g + = 0 d i b gb g = + 5 6 = 0 + 6 = 0 = 0 = 0 or or + 6 = 0 = 6 Always factor the common factor out first Factor or = 0 or or = Factor out the common factor of = 0 = This question is not quadratic but its method of solution is similar. Its properties will be eamined later. Module
0 Section, Lesson Principles of Mathematics e) 3 6 = 0 Solution ( ) 3 = 0 ( )( ) + = 0 = 0 + = 0 = 0 = = 0 = Eample The sum of the squares of two consecutive, odd integers is 7. Find the integers. Solution: Let be the value of the lesser integer. The greater would, then, be +. + ( + ) = 7 + + + = 7 (epanding ( + ) ) + 70 = 0 (simplifying and putting in general form) + 35 = 0 (dividing both sides by ) ( + 7 )( 5 ) = 0 (trinomial factoring) Therefore = 7 (in which case + = 7 + = 5). or = 5 (in which case + = 5 + = 7). So the two integers are 7 and 5 or 5 and 7! In the activity which follows you will get lots of practice on quadratic word problems. Be sure to define your variables carefully and draw a diagram if the unknowns are lengths. With that in mind please move on to the self-marking activity and be sure to check your answers with the answer key which follows Module when it is complete. Module
Principles of Mathematics Section, Lesson Self-Marking Activity. Solve these equations by factoring. Check your solutions. a) = 0 b) 0 = 0 c) + 3 = 0 d) + 9 + 8 = 0 e) + 3 = 0. Rearrange each of the following equations and solve by factoring, if possible. Check your roots. a) 0 = 7 + b) 5 + = 5 c) 3 b g b + g+ 5= 0 d) 9 + = 0 e) + 9= 0 f) 6 6 = 0 g) 3 + 6 + 0 = 0 h) + 8 = 0 3. One leg of a right triangle is 7 m longer than the other leg. The hypotenuse is 7 m long. Find the length of each leg.. Two square checkerboards together have an area of 69 cm. One has sides that are 7 cm longer than the other. Find the length of the sides of each. 5. Find 3 consecutive odd integers such that the product of the second and the third is 63. 6. Two positive numbers differ by and the sum of their squares is 36. Find the numbers. 7. A concrete pathway metres wide is being built around a 0-m 30-m garden. The area of the pathway is 98 m. Find the width of the pathway. Module
Section, Lesson Principles of Mathematics 8. A rectangular piece of card is 5 cm longer than it is wide. A 3-cm by 3-cm square is cut out of each corner, and the four sides are folded up to form an open bo with a volume of 50 cm 3. What were the length and width of the original piece of cardboard? Check your answers in the Module Answer Key. Module
Principles of Mathematics Section, Lesson 3 3 Lesson 3 Solving Quadratic Equations by Completing the Square and by the Quadratic Formula Outcomes When you complete this lesson, you will be able to solve a quadratic equation by completing the square solve a quadratic equation by using the quadratic formula recognize and solve equations in quadratic form Overview Some quadratic equations are easier to solve by factoring than by graphing. However, not all quadratic equations are factorable. One technique used to solve quadratic equations is called completing the square. It follows a similar pattern to what you did in Module, Section to complete the square. When completing the square to solve a quadratic equation you must preserve the equality. When you add a constant to one side of an equation, be sure to add the same constant to the other side of the equation. There are two cases regarding the solution to a + b + c = 0 by completing the square.. When the leading coefficient is. Eample Solve by completing the square: + = 0 Solution + = 0 = ( ) ( ) + = + ( ) = + ( ) = = ± = ± Subtract from each side to isolate the terms with variables Add the square of to each side Simplify the coefficient of Take the square root of each side Add to each side Module
Section, Lesson 3 Principles of Mathematics Remember: + and are the roots of the quadratic equation + and are the zeros of the function + and are the -intercepts of the graph e+, 0j and e, 0j are the points where the graph crosses the -ais. When the leading coefficient is not. If the leading coefficient of the quadratic is not, you divide both sides of the equation by the coefficient before you complete the square. Eample Solve by completing the square: 6 7 = 0 Solution 6 7 = 0 3 3+ These are the eact roots of the equation. Using substitution to check solutions that have fractions and square roots is cumbersome. If you have a graphing calculator, you can use it to do a very efficient check. For instance: 3 3 = 7 F I 7 HG K J F = HG F 3I HG K J = + F I HG K J = 3 + Add seven to each side and divide by 9 3 3 Take the square root 3 3 = ± 3 3 = ± Simplify 3 3 3 + 39. 09. I K J Add the square of Simplify ( 3) to each side Module
Principles of Mathematics Section, Lesson 3 5 On the graph of 6 7 = 0, 3.9 and 0.9 are the approimate -intercepts. To derive a general formula for solving quadratic equations, you can complete the square for the general quadratic equation a + b + c = 0, a 0. a + b + c = 0 b a c + + = 0 a b a c + = a b b + = ± a b F a b c b + + H G I a K J = F + a H G I a K J F b I b + HG a K J = a F b I + HG a K J = c a ac a b b = ± a Standard form Divide by a Isolate terms with on one side ac a b b ac = ± a Complete the square by adding b b a Simplify Simplify to each side Take square root of each side b ac Subtract a Simplify from each side When solving an equation using the quadratic formula, it is important to remember that b± b ac is all divided by a. Think of the results of the quadratic formula as two fractions. g b a b = + a b = a b ac a b ac a b = a b ac a line of symmetry Module
6 Section, Lesson 3 Principles of Mathematics b Recall that, the first term of the fractional epressions, is a the -coordinate of the verte (Module, Section ). So the Ais of b Symmetry is =. The two fractional solutions are values a on the -ais, on opposite sides of the Ais of Symmetry, b ac units away. a You can use the quadratic formula to solve any quadratic equation. When using the quadratic formula, you must set up the equation in the general form of a + b + c = 0, a 0, so that a, b, and c can be correctly identified. Eample Using the quadratic formula, solve + 5 3 = 0. a =, b = 5, c = 3 Solution ± = b b ac a ()( ) () 5 5 3 ± = = 5± 5+ 5± 9 = 5± 7 = = 5 + 7 = 5 7 or = or = 3 Note: This equation could have been solved by factoring. Module
Principles of Mathematics Section, Lesson 3 7 Eample Solve + 3 9 = 0. Solution a =, b = 3, c = 9 b b ac = ± a afa f a f = ± 3 3 9 = 3 ± 9 + 36 = 3 ± 5 = 3 ± 3 5 = 3+ 3 5 = 3 3 5 or Note: This equation does not factor. Therefore, either the quadratic formula or the method of completing the square would have to be used. There are equations that at first glance do not seem to be quadratic. But a closer look reveals that they actually have a quadratic form or pattern. Eample 3 Solve: 5 + = 0 Solution Notice that the equation can be written in quadratic form: ( ) 5( ) + = 0 Temporarily replace by another letter p to obtain a quadratic equation: p 5p + = 0 Module
8 Section, Lesson 3 Principles of Mathematics Factoring, you get Replacing p by you get = or = =± or =± =± (rationalized) The answers are:,,, and Answers should be checked in the original equation. Check: = 5 + = 0 8 0+ = 0 = also works because powers involved are even = e j F HG a fa f p p = 0 p= or p= e j Yes I KJ F H G I 5 K J + = 0 5 + = 0 5 + = 0 Yes, it works = also works because powers involved are even Module
Principles of Mathematics Section, Lesson 3 9 Eample Solve: 5 F I HG 3 K J + 8 F I HG 3 K J + = 3 0 Solution This displays a quadratic pattern. If the equation as p = 3, you can rewrite 5p + 8p+ 3= 0 b gb g 5p+ 3 p+ = 0 3 p = or p = 5 Replace p by a 3 3 = 5 or 3 = f 3 3 = 0 3 + 9 = 0 3 = 3 = 3 a 3 = + 3= = = Check to see that roots work in the original equation. f As always we check our solutions in the original equation. This will be left as an eercise for you. Module
30 Section, Lesson 3 Principles of Mathematics Self-Marking Activity. For each quadratic equation below, state the values of a, b, and c. a) 5 0 b) 3 5 0 = + = ( ) c) 5 3 8 d) 0 = = e) 5 9 f) 9 = =. Solve these equations using the quadratic formula. Be sure to state the formula before substituting values into it. a) + 5 = 0 b) w 3w+ = 0 c) 7w 3w= 0 d) = 5 3. Use the quadratic formula to find the roots of each equation below. a) 3 6 5 = 0 b) = 0 c) 9 8 7 = 0 d) 3 = 0. Find the zeros of the function f defined by a) f: 5 3 b) f = + 6 af e) 0. 0.06 = 0 f) 7 = 0 5. Find the roots of the quadratic equation 3 5 = 0 to one decimal place. 6. Find the roots of the quadratic equation 6 + 5 6 = 0 a) using the quadratic formula b) by factoring Which method do you prefer and why? 7. Find the zeros of the quadratic function f() = + 8 + 5 by a) factoring b) using the quadratic formula c) graphing Module
Principles of Mathematics Section, Lesson 3 3 8. If possible, for each of the following, write an equivalent equation in the pattern a + b + c = 0 for some real numbers a, b, and c, and for some epression involving. ( ) ( ) a) 5 + 3 + 3 + 3 + 8 = 0 b) 3 + 3 3 = 3 ( ) c) 7 = 6 6 d) 6 + = 6 e) + 3 + = 0 9. Find the real number solution set for ( ) ( ) a) + + 3 = 0 b) + + 3 = 0 c) = 6 d) 3 + = 0 Check your answers in the Module Answer Key. Module
3 Section, Lesson 3 Principles of Mathematics Notes Module
Principles of Mathematics Section, Lesson 33 Lesson The Nature of Roots Outcome When you complete this lesson, you will be able to use the discriminant of a quadratic equation to describe the nature of its roots Overview You have found the roots of a quadratic equation by graphing, by factoring, by completing the square and by the quadratic formula. You have obtained a variety of answers some of which were real numbers and others being imaginary numbers. You have found that any quadratic equation of the form a + b + c = 0 can be solved using the quadratic formula The epression b ac is called the discriminant and it enables you to observe the nature of the roots without actually finding the roots. In this lesson, you will consider the value of the discriminant of a given quadratic equation, the roots of the quadratic equation, a description of the roots, and the graph of the corresponding quadratic function. Eample When b ac > 0 is a number that is a perfect square. Solution Equation: 6 + 5 = 0 Value of Discriminant: b b ac = ± a b ac a =, b = 6, c = 5 ( 6) ()(5) = 36 0 = 6 Note that the value of b ac is a perfect square. Module
3 Section, Lesson Principles of Mathematics Roots of Quadratic: r r b g 6 6 6 = + = + + = 5 b g 6 6 6 = = = Description of Roots: r and r are referred to as two real rational roots Graph of the Quadratic Function: y = 6 + 5 y zero zero 3 5 Description of Zeros: The zeros are referred to as two real rational zeros at and 5. Eample When b ac > 0 and its value is a non-perfect square. Solution Equation: 6 + 7 = 0 Value of Discriminant: b ac a =, b = 6, c = 7 ( 6) ()(7) = 36 8 = 8 Note that the value of b ac is a non-perfect square. Module
Principles of Mathematics Section, Lesson 35 Roots of Quadratic: ( ) 6 + 36 8 6+ 8 6+ r = = = = 3+ ( ) () 6 36 8 6 8 6 r = = = = 3 () Description of Roots: r and r are referred to as two real irrational roots. Graph of the Quadratic Function: y = 6 + 7 y e3, 0j 3, b e g j 3+, 0 Description of Zeros: Two real irrational zeros at 3+ and 3 Eample 3 When b ac = 0. Solution Equation: 6 + 9 = 0 Value of Discriminant: b ac a =, b = 6, c = 9 ( 6) ()(9) = 36 36 = 0 Note that the value of b ac is a perfect square. Module
36 Section, Lesson Principles of Mathematics Roots of Quadratic: Description of Roots: r r b g 6 36 36 6 = + = = 3 b g 6 36 36 6 = = = 3 r and r are the same real root. r and r are referred to as double roots Graph of the Quadratic Function: y = 6 + 9 y b g bg (3, 0) Description of Zeros: It has one zero (the graph is tangent to the -ais). Eample When b ac < 0. Solution Equation: 6 + = 0 Value of Discriminant: b ac a =, b = 6, c = ( 6) ()() = 36 56 = 0 Module
Principles of Mathematics Section, Lesson 37 Roots of Quadratic: ( ) 6 + 36 56 + 6+ 0 r = = r ( ) 6 36 56 6 0 = = Description of Roots: r and r are not real roots. Graph of the Quadratic Function: y = 6 + y (3, 5) Description of Zeros: There are no real zeros since the graph does not cross or touch the -ais. Module
38 Section, Lesson Principles of Mathematics Eample 5 For what values of m will + 8 m = 0 have real and unequal roots. Solution For real and unequal roots, b ac > 0. a =, b = 8, c = m The value m must be greater than 6 for real and unequal roots. You were not asked for an equation but an eample that satisfies the restriction, which would be + 8 = 0. Eample 6 Write a quadratic equation that has two different real solutions. Solution A quadratic equation in standard form has two real number solutions when b ac > 0. Choose any values of a and c, then find a value of b that satisfies this inequality. If a = 3, c = b b ()( ) ac > 0 3 > 0 b 3 > 0 b > 3 bgb g 8 m > 0 6 + m > 0 m > 6 m > 6 One possible value of b would be because > 3. Either b > 3 or b < 3. One possible equation is 3 + + = 0. Module
Principles of Mathematics Section, Lesson 39 Self-Marking Activity. State whether each statement is true or false. If the statement is false, rewrite it so it is true. a) The discriminant of + 5 + 6 is 3. b) For each quadratic equation a + b + c = 0 if b ac > 0, then the equation has two real roots. c) For each quadratic equation a + b + c = 0 if b ac < 0, then the equation has no real number solutions. d) 6 + 3 0 = 3 + 8 5 e) The equation + + = 0 has two distinct real roots.. If the discriminant of a quadratic equation has the given value, state the characteristics of the roots. a) 5 b) 5 c) 9 d) 0 e) 50 3. How many times would the graph of y = a + b + c (with a, b, and c as real numbers) intersect the -ais if the value of the discriminant is a) negative? b) zero? c) positive?. Determine the nature of the roots by calculating the discriminant for each equation. a) 8 + 6 = 0 b) a + a + 7 = 0 c) p 6 = 0 d) + = 5 5. Determine the characteristics of the roots of the following equations: a) + + = 0 b) 3 = 0 c) 3 = d) 6 + = 0 e) d i + 9 = 0 6. Given 3 m + 3 = 0, for what values of m would the roots not be real? Module
0 Section, Lesson Principles of Mathematics 7. Find value(s) of k so that each equation has one real root. a) k 6 + = 0 b) + (k 8) + 9 = 0 8. For what values of k will the equation + + ( k k ) = 0 have eactly one root? 9. State the nature of the roots of a + b + c = 0 given the graph of y = a + b + c: a) y b) c) y 5 y 5 5 5 5 5 5 5 0. Sketch the graph of a quadratic function with no real zeros.. Sketch the graph of a quadratic function whose discriminant is positive and whose verte is in the second quadrant.. Find the value of k for each of the following situations: a) 3 + k = 0 has two real roots b) + k + (k + ) = 0 has a double root c) k + 8 = has no real roots Check your answers in the Module Answer Key. Module
Principles of Mathematics Section, Lesson 5 Lesson 5 Radical Equations Outcome When you complete this lesson, you will be able to solve radical equations Overview A radical equation is one that contains radicals or rational eponents. To solve radical equations you try to eliminate the radicals and obtain a linear or quadratic equation, which you have already had eperience in solving. The following property plays a key role in the simplification process. If a = b, then a n = b n. It means that both sides of an equation can be raised to the same power as shown in the following eamples. Remember: Square roots and all other even roots are only defined if the radicand (the epression under the radical sign) is nonnegative. For eample: Eample Solve: = Solution = e j = = 6 Check: is defined only for 0 5 is defined only for 5 0 or 5 6 = Square each side to remove the square root sign. Remember that = and if you square, you get. Module
Section, Lesson 5 Principles of Mathematics Eample 3 Solve: Solution: = 0 Before raising both sides of an equation to the nth power, you must isolate the radical epression on one side of the equation. 3 = 3 3 3 e j = = 6 Rearrange Cube each side remembering that 3 and e j 3 3 3 = = = 3 3 Check: 3 6 = 0 = 0 The following is an equation containing an eponent. Eample 3 Solve: 3 = 8 Solution 3 = 8 3 3 3 e j = 8 3 = e 8j = = Raise both sides to the power Simplify 3 Check: e j 3 = 8 3 = 8 3 = 8 Module
Principles of Mathematics Section, Lesson 5 3 Raising both sides of an equation to the nth power may introduce etraneous or false solutions. So when you use the procedure, it is critical that you check each solution in the original equation. For instance, squaring both sides of the equation. The result is etraneous. The original equation has no real number solution. Eample = produces = Solve: + 3 = 6 Solution 3 = 6 e 3j = b6 g = or = Check: 3 = 36 + = 36 + 6 + 8 = 0 = 0 + 3 = 6 b + = 6 6 = 6 Answer: = g b gb g + 3 = 6 + 3 = 6 Isolate the radical Square both sides Rearrange 8 6 e j b g When you squared both sides of the equation 3 = 6 you introduced a second equation, that being 3 = b6 g. Thus etraneous roots may be introduced and a check of the roots is always necessary. Module
Section, Lesson 5 Principles of Mathematics If a radical equation has more than one term with a variable in the radicand, you may have to isolate and raise it to a power more than once as shown in the following eample. Eample 5 Solve: + 3 + = Solution + 3 = + + ( + 3) = ( + + ) = ( + + ) ( + + ) ( + ) = ( + ) ( )( ) ( ) ( ) = + + + + + + + 3 = + + + + + = + + + = + + + = + 3 = 0 3 + = 0 = 3or Rearrange Square each side Simplify Rearrange Square again Check: bg 3 + 3 3+ = b g 3 = + 3 + = 0 = The solution set is {3, }. Module
Principles of Mathematics Section, Lesson 5 5 Self-Marking Activity. State whether each statement is true or false. If the statement is false, rewrite it so that it is true. a) For each real number, is a real number if and only if 0. b) For each real, 3 is a real number if and only if 0. c) For each real, d) The real number solution set of e) The equation ( ) = is equivalent to the sentence ( ) = or =. f) For each real number, if = 6, then = g) Squaring both sides of + = 3 yields + = 9.. Simplify each of the following epressions. ( ) ( + ) ( + ) a) b) 5 c) 5 =. ( ) ( ) { } = 3is 3, 3. 6. 3. Find the real number solution set for each of the following equations. Check your solutions. a) + = + 7 b) = 5 c) + 3 + = d) 3 + = 0 e) = 3 + f) + 6 = g) 3 + = 3 h) + = + Check your answers in the Module Answer Key. Module
6 Section, Lesson 5 Principles of Mathematics Notes Module
Principles of Mathematics Section, Lesson 6 7 Lesson 6 Rational and Absolute Value Equations Outcomes When you complete this lesson, you will be able to solve rational equations solve absolute value equations Overview You have solved linear equations, quadratic equations, radical equations, and non-rational equations. You can solve rational equations by using many of the techniques and properties you have used to solve other equations. An equation involving rational epressions is a rational equation. The algebraic method of solution is based on the least common multiple. The least common multiple (LCM) of two or more polynomials is the simplest polynomial that is a multiple of each of the original polynomials. Eample Find the least common multiple of the following polynomials. You factor each polynomial and use each factor the greatest number of times it occurs in each polynomial. Polynomials Least Common Multiple a) and ( ) (each factor occurs once) b) and = ( occurs twice in first term) c) 9 and 3 ( + 3)( 3) ( 3)( + 3) and ( 3) d) ( + ) and ( )( )( + ) ( )( ) and ( )( + ) The least common denominator (LCD) of two or more fractions is the least common multiple of the denominator of the fractions. Module
8 Section, Lesson 6 Principles of Mathematics To solve a rational equation: if the denominator can be factored, factor it first. look in the denominator for restrictions: values of the variable which are forbidden because they make the denominator equal zero. (Remember, you can t divide by zero.) multiply each term on both sides of the equation by the least common denominator of the terms. simplify and solve the resulting equation. Then check! Eample 3 + = Solution Denominators are fully factored, so choose lowest common denominator to be. The restriction is 0. 3 + = + = 6 = Multiply both sides by Simplify Check in original equation. 3 + = 3 3 = Check is okay. You can also use your graphing calculator if you have one. Press Enter Press Y= y y = + = 3 GRAPH Press nd CALC 5: Intersect Module
Principles of Mathematics Section, Lesson 6 9. Adjust your WINDOW so that X min =, X ma = 3, Y min =, Y ma = 3. Press ENTER to confirm the st Curve. 3. Press ENTER to confirm the nd Curve.. Arrow left or right to place the cursor as close as possible to the point of intersection of the two curves. 5. Press ENTER to confirm your Guess. 6. Record the value at the bottom of the screen. This is your solution,. Point of intersection: (,.5) Eample Solve: 3 6 = + Solution: Denominators do not need factoring. Restricted value in denominator:. Lowest common denominator: F 3 b I g HG K J = b g+ b g 3 = + 6 = F HG 6 I K J Multiply by When you check your answer, is not a permissible value as the denominator becomes 0. There is no solution to the equation. Eample 3 Solve: 3 + = + Solution One denominator needs to be factored to ( + )( ) first. Restricted value:, Lowest common denominator: ( )( + ) Module
50 Section, Lesson 6 Principles of Mathematics 3 I b gb + g HG + K J = b gb + g HG b gb + g KJ + b gb + gbg Check: = or = 5 = 3b g + = + b g 6 = + 6= + 6= 6 = or = 5 F 3 = + + 3 3 = + 3 3 = + 3 0 = 0 + 5 = 0 The last type of equation to be eamined in this lesson is the absolute value equation. The absolute value of a number can be thought of as its distance from 0 on a number line. Any positive number and any negative number are each a positive distance from 0. The absolute value is concerned with distance rather than direction. Definition The absolute value of any real number a, written as a, is defined as a = a if a 0 a = a if a < 0 Eample: = = + b g b gb g d i b gb g ( ) F = 5 35 bg + = 5 5 + 5 = + 6 5 = + 5 5 = I Multiply by LCD Simplify Factor Module
Principles of Mathematics Section, Lesson 6 5 There are two values that have a distance of from 0. They are and. So, if =, then = or =. distance is distance is 0 Eample Solve and graph: 3 = Solution b g 3 = 3 = = 5 Check: = 5 5 3 = = or or b g 3 = 3 = = = 3 = = = The solutions of 3 = are translations of the solutions of =, 3 units to the right. 0 5 or points which are a distance of from 3 distance of 0 3 5 Module
5 Section, Lesson 6 Principles of Mathematics Eample 5 Solve: 3 = + 6 Solution Since the variable epression in the absolute value signs can be positive or negative, you must solve two equations. 3 = + 6 + 6= 0 ( )( ) + 3 = 0 = 3 or + 3 = + 6 + 3 = + 6 d 7 + 6= 0 6 = 0 i b gb g = 6 or Check: b g = 3 b g b gb g 3 3 3 = 3 + 6 9+ 9 = + 6 8 = 8 b g = b g 3 = + 6 = 8+ 6 bg = 6 bg 6 3 6 = 6 + 6 8 8 bg = bg 3 = + 6 = = = 3 or =. Module
Principles of Mathematics Section, Lesson 6 53 Eample 6 A certain number is greater than its reciprocal. Find the number. Solution Let be the number and its reciprocal. = () = ( ) = = 0 So there are two solutions, (.09 appro.). ( ) ( )( ) () ± ± + ± 5 ± 5 5 = = = = + 5 5 (.09 appro.) and 5 5 Checking these two solutions will be left as an eercise for you. Now, for some practice, please complete the following Assignment. Be sure to check your answers when you are finished. Module
5 Section, Lesson 6 Principles of Mathematics Self-marking Activity. State whether each statement is true or false. If the statement is false, rewrite it so it is true. gf I HG K J = gf I HG K J = + a) If, then. 6 b) If 3, then 3 3 + 6. c) If 0, then the equations = 0 and = 0 are equivalent. d) If and then is equivalent to + = 3, 3 + + = + 3. b gb g b gb g b b. Solve each of the following rational equations. Check your solutions. a) = 9 5 b) + = 3 7 7 c) = 3 d) + = + 3 3 + e) f) + + + 3 + 9 + 7 = 0 = 3 3 3 9 3 3 3. Solve the following absolute value equations.. Two positive integers differ by and their reciprocals differ by a) 3 = b) = 7 c) 5 + = 3 d) + = 0 3 e) = f) 5 = 3 + 7 5. Find the integers. 5. The sum of a number and its reciprocal is and the number is greater than. Find the number. Module
Principles of Mathematics Section, Lesson 6 55 Check your answers in the Module Answer Key. Review Module before attempting the review questions beginning on the net page. These questions should help you consolidate your knowledge as you prepare for the Module Section Assignment.. Module
56 Section, Lesson 6 Principles of Mathematics Notes Module
Principles of Mathematics Section, Review 57 Review. Solve the following quadratic equations by factoring: a) 3 = 6 b) 5 = 8 c) 3 3 8 = 0 d) 6 5 + = 0 e) = 0 f) 3( ) ( + ) + 5 = 0. Find the roots of the equation by graphing: a) 8 = 0 b) 8 = 5 c) = 0 3. Solve by completing the square: a) + 6 = 0 b) 3 8 = 0. Solve by using the quadratic formula: a) 7 = 0 b) + + = 0 5. Without solving the equation, determine the nature of the roots of: a) b) c) d) + 9= 0 3 = d i 3 = 0 6 + = 0 6. Find the values of k so that the equation, 3 + 6 + (3 k k ) = 0, has real and equal roots. 7. For what values of k will the equation 3 + k + = 0 have two different real roots? Module
58 Section, Review Principles of Mathematics 8. a) For what values of k will the equation + 5 = k have no real roots? b) What does this tell you about the graph of the function for those values of k? 9. What is the discriminant for 7 + 5 = 0? 0. Solve the following equations: a) b) + 3 + + = + 5 = c) d) e) f) g) 3 3 = 5 6 + 3 + = + d i d i + + + = 8 = 0 + 5 = 3. The product of two consecutive positive integers is 7. Find the integers.. If (6 ),(3 ), and ( ) are the length of the sides of a right triangle, find the value of, if ( ) is the hypotenuse. 3. Two positive numbers differ by. The sum of their reciprocals is /3. Find the numbers. Check your answers in the Module Answer Key. Now do the Section Assignment which follows this section. Module
Principles of Mathematics Section Assignment. 6 PRINCIPLES OF MATHEMATICS Section Assignment. Module
6 Section Assignment. Principles of Mathematics Module
Principles of Mathematics Section Assignment. 63 Total Value: 50 marks (Mark values in brackets) Section Assignment. Algebra Answer the following questions in the space provided. To receive full marks, solutions should be legible, well-presented, and mathematically correct. (). Solve by factoring: 3 = 0 (). Find the roots of the following questions by using the quadratic formula. a) + = 0 Version 0 Module
6 Section Assignment. Principles of Mathematics b) = 7 (3) 3. Solve by completing the square: + 6 + = 0 (3) Module Version 0
Principles of Mathematics Section Assignment. 65 (3). Solve by any method. a) 6 5 = (3) b) 3 + 3 = 7 () c) 3 3 + = 0 Version 0 Module
66 Section Assignment. Principles of Mathematics 5. Solve for : a) + 5 = 0 () d i d i b) + + + = (5) Module Version 0
Principles of Mathematics Section Assignment. 67 (5) c) + = + 3 3 3 b g () d) + 6 = 7 Version 0 Module
68 Section Assignment. Principles of Mathematics 6. a) Give the discriminant for the quadratic equation, + 3 + = 0 () b) State the nature of the roots. 7. If the discriminant of the quadratic equation, 3 + c = 0, is 5, find a) the value of c () b) without solving for the roots, state the nature of the roots. () Module Version 0
Principles of Mathematics Section Assignment. 69 (3) 8. Find the values of k so that the equation + 6 + k = has two different real roots. (3) 9. The sum of a number and 5 times its reciprocal is equal to 8. Find the numbers. Version 0 Module
70 Section Assignment. Principles of Mathematics 0. How wide a uniform border should be left on a plot of land 7 m by m to have a rectangular area of 5 m available for flowers. () (Total: 50) Module Version 0