MATH 5a SOLUTIONS TO REVIEW SHEET FOR EXAM I The first Math 5a midterm will be Friday, February 9th from 4 p.m. Location: Goldsmith 00 The exam will cover Sections..5: Section.: Real Numbers Section.: Exponents and Radicals Section.: Algebraic Expressions Section.4: Rational Expressions Section.5: Equations No calculators will be allowed on the exam. Our exams are closed book: no notes, books, calculators, cell phones or internetconnected devices will be allowed. On the exam, you will be asked to show all your work. We are grading your ability to clearly communicate your understanding of how to solve each problem much more than your ability to get the correct answer. In fact, a correct answer with little or no work might not receive any points at all. Showing your work clearly also gives us justification for assigning partial credit. IMPORTANT: This review sheet is much longer than the exam. It is similar to the exam, however, in that it contains a mixture of straightforward, intermediate and challenging problems. Remember that there are self-quizzes (with solutions) available on LATTE that you can also use to test your knowledge. OVER for SOLUTIONS
Practice Exam SOLUTIONS (Problems - 4). xy 75xy xy(y 5) xy(y 5)(y + 5). (a) A [, π) and B [, ) (b) A B [, ) (c) A B [, π) (d) Z A {,, 0,,, }. 6 ( 4 ) 4 6 6 64 4. Let p(x) 5x x + 4x + π. (a) We write the polynomial with highest degree terms first: 4x x + 5x + π. So the degree of the polynomial is. (b) We first simplify the polynomial as much as we can. Our polynomial has two linear terms that we can subtract to obtain 4x + 9/x + π. So the number of terms is. (c) The constant term is the term that does not have the variable x in it. So π is the constant term of p(x). (d) In part (b), we added the linear terms together to obtain 9 x. So the coefficient is 9. 5. We exclude all values that make the expression x undefined. This happens when x x and also when the x < 0 since we cannot take square roots of negative numbers. So the domain of the expression is: {x : 0 x < or x > }, or equivalently [0, ) (, ). 6. rs s r r s s r r/ s s / r / r/ / s ( /) r /6 s 4/. 7. (a b) (a b ) (a ab + b )(a b ) a + b a + b + + a b + a 7 b a 4 b + a b 4.. x 5 is not a solution to (x 5) 0 since it s not even in the domain of the expression x 5 on the left. And even if we clear the denominator, it would not work since we clear assuming that x 5 is not in the domain. We are still excluding the value of x 5 at the second step when we solve x 5 0 for x. 9. Simplify ( ) x y n z m (a) ( x ( ) y n (n+) z m m) ( x 4 y z m) x y 6 z m x y n+ z m y 6 x z m (b) x+ x (x+) (x+) x x (x+) x (x + ) x + 4 (c) x5 x 6x x x 0 x(x4 x 6) 5)(x + ) x + 7x + x x (x + )(x + ) (x x(x ) (x 5)(x + ) (x + )(x 5) x(x ) x + x(x )(x + ) (x + )(x + )
0. Solve (a) x x x x x(x ) x(x ) (We multiply by x(x )) x x x 9x 0 4x x + Using the quadratic formula we obtain: x ± 44 4 ± 4 6 ± 6 (b) x 4x x 4x 0 x(x 4) 0 x(x + )(x ) 0 So the solutions are x 0, x, x. (c) x4 6 x + 0 (x + 4)(x 4) 0 x + (x + 4)(x + )(x ) 0 ± 96 9 (x + ) Even though x is a root of the numerator, it is not in the domain of the expression. So we exclude x. Moreover, the discriminant of x + 4 is 6. So the only solution to the equation above is x.. Given L L 0 L L 0 L L 0 v c v c L L 0 v c ( L v c v, we solve the equation for speed v. c v c (we divide both sides by L 0) (square both sides) (isolate v /c ) L 0 L L 0 ) (multiply through by c ) (finally, take the positive square root since speed is non-negative). We are given the following equation for capacitance + C t C C where C t is the total capacitance, and C and C are the capacitances of two capacitors. C t 0µF and the capacitance of one of the capacitors is 00µF. Let s choose C 00µF, without loss of generality. Then, 0 00 + C 0C C + 00 (multiply through by 00C ) 9C 00 C 00 9 µf.
. The total amount of snow A (in feet) at a given time t (hour) is given by the equation: A(t).5t t +. We solve t in terms of A. ta + A.5t (multiply through by t + ) ta.5t A t(a.5) A t A A.5. 4. True or False (a) False. B is the set of real numbers strictly greater than and strictly less than. So A B. (b) True. Let s look at the discriminant of the denominator x x+5. We compute it to be: 4 4()(5) 56. Since the discriminant is negative, there are no real solutions to the equation x x + 5 0. In other words, the expression is defined for all reals. (c) True. The monomial x is always a factor of a polynomial without constant term. Thus, x 0 is a root of such a polynomial. OVER for Practice Exam SOLUTIONS
Practice Exam SOLUTIONS (Problems 5 - ) 5. Factor the following. (a) x 64 x 64 (x )(x + ) (b) 6x x 6x x (x + )(x ) (c) 50 50 5 4 5 (5 ) 6. Consider the polynomial x + 7x x 4 + 6 (x + ) (a) What is the degree of the polynomial? degree of polynomial 4 (b) How many terms does it have? Be careful! First we need to reduce the polynomial: x + 7x x 4 + 6 (x + ) x + 7x x 4 + 6 x 6 7x x 4 We can see the polynomial has terms. (c) What is the constant coefficient? version) 7. Evaluate the following: (a) 9 (b) 9 9 ( 9) 7 64 4 or ( ) () 4 constant coefficient 0 (also from reduced (c) 4 54 4 54 4 6 9 6 6 6
. Simplify the following expressions: (a) x+ x x x+ x x [ x x + ] x [ x x x(x + ) x + ] x(x + ) [ ] x x x x(x + ) [ ] x x(x + ) x(x + )(x ) (b) 5 5 [ ] 5 [ 5 ] 40 40 [ ] 5 40 40 ] [ 40 9. Let A {x : x 4x x 0}, B {x : x > }, C {x : x } (a) Draw a number line and mark the points in A on your number line. First we need to figure out what x values are in A, so we solve: On a number line, this looks like: x 4x x 0 x(x 4x ) 0 x(x 6)(x + ) 0 so x 0, x 6 or x so A {, 0, 6} - - - 0 4 5 6 7 (b) Find A B. A B is the set of elements that are in A that are also in B, so A B {0, 6}.
(c) Find A C. Draw your answer on a number line. A C is the set of elements in A together with the elements in C, so that means all the numbers less than or equal to and also the number 6. On a number line, it looks like this, with A C in red: - - - 0 4 5 6 7 0. Find the domain of the expression x + 5. Write your answer in interval notation. We can t divide by zero, so we need x 5. We also need the number inside the square root to be greater than or equal to 0, i.e., x + 5 0 so we need x 5. Taking these two restrictions together makes the domain {x x > 5}. In interval notation: ( 5, ).. Simplify the following expressions: (a) x + 7x x x 0 x + 5 x x + 7x x x 0 x + 5 x(x + 7) x (x + 5)(x 4) x + 5 x x(x + 7) (x + 5)(x 4) x + 5 x x + 7 x (x 4) (b) ( a / b / ) ( a / b / ) ( b / a / b / a /9 b/ a /9 ). The area of a particular shape can be represented by the equation A πr r Solve the equation for r. A πr r A( r) πr A Ar πr A πr + Ar A r(π + A) A π + A r
. Solve the following equations: (a) x x 6 x x 0 x x 6 x x 0 x x 6 0 (x )(x + ) 0 so x or x Since x is a zero of the denominator, it is not in the domain of the expression x x 6 and therefore cannot be a solution. So x is the only solution. x x (b) 6x x 4 6x x 4 6x x 5 0 (x + 5)(x 7) 0 so x 5 or x 7 You could also use the quadratic formula to find the zeros. 4. A hunting seagull swoops down from the sky to catch fish underwater. Suppose the height of the seagull above sea level is given by h t 4t + 40, where t is the amount of time, in seconds, after the seagull begins its swoop down toward the water. How long after the seagull begins its swoop does it come out from under the water? We need to solve t 4t + 40 0 (t 0)(t 4) 0 so t 0 or t 4 (You could also use the quadratic formula to find the zeros.) Since we want to know when the seagull comes back out of the water, we need the second time that the seagull is at sea level, i.e., 4 seconds after it begins its swoop. 5. Use the discriminant to find out how many solutions there are to the equation D b 4ac 6 4()() 6 60 x + 6x + 0. Since the discriminant is 0, the equation has exactly one solution.
6. A student tries to solve the following equation. Explain what is wrong with their solution. x 4 x 4 (x )(x + ) 4 (x ) x + 4 x The student s work is ok, except that they need to check whether their answer is in the domain. It turns out that if you plug x into x 4, you get 0 in the denominator, so x the expression is undefined (and, therefore, not equal to 4). So x is not a solution to the equation. 7. A canoeist heads upriver at a speed of x km/h. She travels up the river for km and then back down the river for km, and the current of the river flows at a steady km/h. The following equation represents the fact that she spends a total of 4 hours canoeing on the river. x + x + 4 (a) How fast was she canoeing? We need to solve x + x + 4 [ (x )(x + ) x + ] 4(x )(x + ) x + (x )(x + ) + (x ) (x + ) 4(x )(x + ) x x + (x + ) + (x ) 4(x ) x + + x 4x 4 0 4x 6x 4 0 x x 0 (x + )(x ) so x or x The canoeist is not traveling at negative speed, so she travels at km/h. (b) Notice that if the canoeist goes upriver at km/h, her efforts will be canceled out by the speed of the current and she will not be able to travel up the river. Explain how this fact is reflected in the equation. x is not in the domain of + so the equation also tells us that this x x+ speed is not a possible answer.. Determine whether each of the following three statements is true or false, and explain why the statement is true or false. Note: In mathematics, true means that the statement must always be true. False means that the statement may sometime be false.
(a) The equation x has no real solutions. FALSE: x is a solution since ( ). (b) The equation x 0 0 has no solutions. FALSE: x is a solution since 0 0. (In fact, any real number is a solution.) (c) The equation x 0 has no solutions. TRUE: Since x 0 0 for any x, there are no numbers x such that the product of x and 0 is. Note: The in b and c indicates multiplication. OVER for Practice Problems Solutions
9. (a) [ 5, ) Practice Problems SOLUTIONS (Problems 9-5) (b) (, ) (c) (, + ) 0. (a) A B (, + ) (b) A B (4, 5]. Remember that the set of integers is the set...,,,, 0,,,,.... The rational numbers are all numbers of the form p, where p and q are integers (with q 0). Any q integer is also a rational number since it can be written as itself. Therefore: (a) S Z { 0,, 5 } (b) S Q { 0,, 0.4,, 5 } 5 Note that 0.4 4, so it s rational number. 0. (a) (b) ( x y z 4 x y z ) x 4 y 6 z x y 4 z 6 x y z (a 4 b c) (a b c ) a b 6 c a 9 b 6 c a c (c) m+n ( m + n ) m+n m + m+n n m+n m + m+n n n + m. 4 6 5 4. (a) x y ( ) (x y) ( ) xy x y 4 (b) 4 x x x 4 x x ( 4 + ) x (c) x 9x x x (9x) x 5. (a) 9 9 ( ) (b) 9 9 (c) ( 7) 4 ( 7) 4 x x x x( 6 + ) x x 7 x ( 7) 4 ( ) 4 x ( 7 ) x 7 6 6. (a) (x 5 y) ( x y ) x( 5 + ) y (+ ) x 9 0 y 5 (b) (6x 5 4 y ) 5 6x (x 9 y) x y y 6x 9 6 y 4 7. The polynomial is x5 x7 + x 5. (a) The degree of the polynomial is 7.
(b) The coefficient of the term of highest degree is. (c) The constant term is 5. (d) The coefficient of the term of degree is, the coefficient of the term of degree 5, the coefficient of the term of degree 7 is. is. ( x 5 + 5 x + x 7 ) (4 x x ) x 5 x + 7 x + 9. (a) ( y 9 y + 9) (y + ) y 9 + y 7 y y + 9y + y 9 0 7 y + y + (b) (a b)(a + b)(a + b ) (a b )(a + b ) a 4 b 4 40. (a) 9x + x + 4 (x + ) (b) 6x 7x 4x x(6x 7x 4) x(x + )(x 7) (c) x y z yz 5 yz (x y 9z ) yz (xy z)(xy + z) (d) x (x 4 )(x 4 +) (x )(x +)(x 4 +) (x )(x+)(x +)(x 4 +) ( ) ) (e) x x5 00 x x4 x ( )( x + x 00 0 0 (f) t 4 + t (4t )(t + ) (t )(t + )(t + ) (g) x n x n 4 (x n x n ) (x n 4)(x n + ) 7x 49 4. (a) To find the domain of 4x 9x, find the zeros of the denominator: 4x 9x 0 x(4x 9) 0 x(x )(x + ) 0 x 0, x ±. So the domain is all real numbers except x 0, x and x. In interval notation, this is (, ) (, 0) (0, ) (, + ). (b) To find the domain of x, note that x must be greater than or equal to zero. So the domain is the set { x : x }. In interval notation this is [ + ). (c) The domain of x is the set { x : x > 0 }. In interval notation this is (0, + ). x (Note that x 0 is not in the domain, because it would make the denominator zero.) x (d) To find the domain of, first note that since there is an odd root in the 6 x numerator we only need to look at the denominator. From the denominator, we see that x cannot equal 6. So the domain is { x : x 6 }. In interval notation, this is (, 6) (6, + ). 4. (a) Problem # : (x + )(x + ) (x + 5)(x ) x + x + x + x 5 x + 6x + 5 x 7x + x + x + x + x 5 x 7x + x + 6x + 5 (x )(x ) (x + 5)(x + ) (x + )(x ) (x + 5) 4x (x + 5) (b) Problem # 9: + x (x + ) + x x + 6 x + x + x +. (c) Problem # 4: x + x + x + (x + )(x + ) x + (x + )(x + ) x + x (x + )(x + ) (x + )(x + ).
4. (d) Problem # 46: (e) Problem # 60: + v v 44. (a) a + h a h v + v v v 5 x 5(x ) (x ) (x ) + c c a (a + h) a(a + h) h (c ) + c (c ) c v v v + v v v v + v h a(a + h) h c c c c 0x (x ) c c c c h a(a + h) h a(a + h) c c (b) (5 + h) 5 h 5 (5 + h) 5 (5 + h) h 0h h 5 (5 + h) h 5 (5 + 0h + h ) 5 (5 + h) h 0h h 5 (5 + h) h 0 h 5 (5 + h). It s OK to leave the answer in this form. You can also write it as 5 5 0h h 5 (5 + h) h 0 h 65 + 50h + 5h. 45. (a) Problem # : R +. We multiply both sides by the LCD, which is R R RR R, getting R R RR + RR R R RR RR R (R R) RR R RR R R. ax + b (b) Problem # : ax + b (cx + d) ax + b cx + d. To cx + d solve for x, collect all the terms with x on one side of the equation and collect all the other terms on the other, getting ax cx d b. Now isolate x by factoring x out of both terms on the left: x(a c) d b. Finally, solve for x by dividing both sides by a c: x d b a c. (c) Problem # 7: V πr h V πr h r V V πh r ± πh. (d) Problem # 70: For x 6x + 0 we have a, b 6, c in the quadratic formula, yielding: x 6 ± x 6 ± 4 x ±.
(e) Problem # 76: For + 5z + z 0 we have a, b 5, c in the quadratic formula, yielding: z 5 ±. (f) Problem # : Calculating D 5 5 4 4 < 0, we see that the equation has no real solutions. 46. P eσat 4 T 4 P P eσa T ± 4 eσa L 47. T π T L π T 4π L L T π. 4. The following is a quadratic equation with solutions 4 and 5: (x 4)(x + 5) 0. We can multiply this out and write it in the usual form: x + x 0 0. Notice that if we multiply both sides of this quadratic equation by any non-zero number, we get another quadratic equation with the same solutions. For example, x +x 40 0 also has solutions x 4 and x 5, since solving x + x 40 0 gives us (x 4)(x + 5) 0 x 4, x 5. Similarly, x + x 60 0 also has solutions x 4 and x 5; so does x x + 0 0, etc. 49. A quadratic equation has exactly one solution precisely when the discriminant is zero. First we must rewrite the equation as x + px + 0. Recall that the discriminant of the equation ax + bx + c 0 is b 4ac. Note that here b p. So we want p 4()() 0 p 6 0 p 6 p ±6. 50. x πx+ 0. The discriminant of this quadratic equation is π 4, which is positive (since π > and hence π > 9, meaning that π 4 > 0). So the equation has two real solutions. 5. (a) x x + 4x 6 x + 6 x 9 x. 6 x (b) x + 0. The LCD is (x )(x ). The result is x x x + x 0 x(x ) + x 0 x 5x 0. Then use the quadratic formula, getting (c) t t x 5 ± 5 + 6 4 x 5 ± 4. 4 0 t 0. So the solution to the equation is t 0.
(d) x 49x x 49x 0 x(x 49) 0 x(x + 7)(x 7) 0 x 0, x ±7. So the solutions are x 0, x 7 and x 7. (e) r 4 + r 4 0 (r + 4)(r ) 0 r + 4 0 or r 0. The first equation has no solutions. The solutions to the second are r ±. (f) x 9 x 0 x 9 0 x 9 x, x. x, x. So the solutions are (g) x x + 5 x x + 5 0 (x )(x 5) 0 x, x 5. But x x makes the denominator of x x + 5 zero, so it can t be a solution to x the equation. So the only solution is x 5. (h) + x x + LCD is 6x. Clearing denominators gives us x+ +x x 6. So the solution is x 6. (i) x + 9 x 0 x + 9 0 x 9. There are no real numbers which, when squared, equal 9. So the solution set is. 5. The discriminant of the equation is p + 4q. Note that the expression p + 4q can never be < 0, since it is the sum of two squares. By hypothesis, neither p nor q can be zero, so p + 4q can t 0. So p + 4q must be > 0. So the equation x px q 0 has two real solutions. 5. In order for the oxygen consumption to be the same for a walker as for a runner, it must be true that 5 t + 5 t + 0 t + 0. Solving this equation for t gives 5 t + 5 t+0 t+0 5t +5t+0 t+0 5t +5t t 5t t 0 t(5t ) 0 t 0 or t 5. Discard t 0 since the polynomial t + 0 models the speed of the runner only for 4 t 9 mph (though of course if neither the runner nor the walker is moving, their oxygen consumption would be equal). So when they are both moving at 5 5.6 mph, their oxygen consumption is the same. 54. Set R 7 in the formula for surface area of the donut, getting π (49 r ). We want the surface area to be 0π, which means solving the equation π (49 r ) 0π for r. We can first divide both sides by π, getting 49 r 0 9 r 0 r 9 r ± 9. But the radius of the donut cannot be a negative number, so the only solution is r 9. 55. The ball s height h (in feet) above the ground at time t (in seconds) after being thrown is given by h 6t + t + 4.
(a) Solve for t when h 0: 0 6t + t + 4 6t t + 6 0 6(t t + ) 0 t t + 0 (t )(t ) 0 (t ) 0 t 0 t. So at t second, the ball reaches a height of 0 feet. (b) Solve for t when h 0: 0 6t +t+4 6t t 4 0 4(4t t ) 0 4t t 0. Solve this quadratic equation using the quadratic formula: t ( ) ± ( ) 4(4)( ) (4) ± 64 + 6 ± 0 ± 4 5. Note that 5 is >, so 4 5 is negative. Since the time value can t be negative, 4 5 is not a valid solution. So the only solution is t + 4 5 seconds. 56. According to the Minkowski model of space-time, the distance D between you and an event in your past or future is D t + x + y + z. If we solve this equation for t, we get D t + x + y + z t x + y + z D t ± x + y + z D. Although it wasn t explicitly stated in the problem, t represents time, which can t be negative. So t x + y + z D. 57. (a) x y x y. False. For example, if x 5 and y 4, then x y 5 6 9, which is not equal to 5 4. Note: You could also point out that (x y) x xy+y, which is not the same as x y. (b) Every rational number is also an integer. False. For example, the rational number is not an integer. (c) Every integer is also a rational number. True. If n is an integer it can be written as n, which is a rational number. (d) 6 is undefined. False. 6 6 6 4. x (e) The polynomial + x + 4 can be simplified by multiplying it by 6 to clear denominators. In other words, x + x + 4 x + x + 4.
False. We cannot simplify the polynomial x + x + 4 by multiplying it by 6, since multiplying by 6 changes the polynomial. To see this, let x. Then whereas So x + x + 4 () + + 4 4 6, x + x + 4 () + () + 4 9. x + x + 4 x + x + 4. (f) The largest real number in the interval [, ) is 7. False. The number 7., which is greater than 7, is in the the interval [, ). In fact, there is no largest real number in the in the interval [, ). (g) The equation 0 has no real solutions. x 64 True. A rational expression equals zero only if the numerator equals zero and the denominator does not. But the numerator of is a constant, so it can x 64 never be zero. 5. No, they do not have the same solution set. If we solve the equation x + 5x + 6, x + we get the following: x + 5x + 6 x + x +5x+6 x+ x +4x+4 0 (x+)(x+) 0 x. But x is not a valid solution since it is a zero of the denominator x + in the equation, so we must throw it out. Hence the equation x + 5x + 6 has no real x + solutions. However, if we solve the equation x + 5x + 6 x + we see that x is in the solution set.