Chapter 3. Algebra. 3.1 Rational expressions BAa1: Reduce to lowest terms


 Malcolm Thompson
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1 Contents 3 Algebra Rational expressions BAa1: Reduce to lowest terms BAa: Add, subtract, multiply, and divide BAa3: Find and simplify products of the form (ax + b)(c/x + d) BAa4: Evaluate BAa5: Simplify complex fractions Factoring quadratics BAb1: Factor a quadratic trinomial in two variables BAb3: Factor a perfect square trinomial in more than one variable BAb4: Recognize relationships among coefficients of a factorable trinomial Operations on Polynomials BAe1: Simplify polynomial expressions involving products of binomials in one or more variables BAe: Factor a difference of squares BAe4: Divide a third degree by a first or second degree to obtain the quotient and remainder BAe5: Raise a binomial to the second or third power Graphs BAc: Find the slope of a line from an equation BAc3: Find the slope of a line from the coordinates of two points BAc4: Find the equation of a line from the coordinates of two points BAc5: Find the equation of a line given 1 point and the slope BAc6: Determine and/or plot the intercepts of a line, given an equation BAc9: Given an equation determine if a point is above or below the graph Quadratic/Rational Equations BAd1: Quadratics of the form ax + b = 0 or (ax + b) + c = BAd: Quadratics of the form ax +bx+c = 0 with a 1 by factoring over the integers
2 3.5.3 BAd3: Quadratics using the quadratic formula, where the discriminant might be positive, negative or zero BAd4: Equations quadratic in u, where u = x,x 3, BAd5: Rational equations leading to quadratic equations BAd6: Rational equations leading to linear equations BAd7: Solve literal equations involving rational expressions Radicals and Fractionial exponents BBa4: Express radicals in terms of fractional exponents and vice versa BBa1: Evaluate numerical expressions involving radicals and/or fractional exponents BBa: Simplify radical expressions BBa3: Rationalize the numerator or denominator of a quotient involving radicals BBa5: Multiply radical expressions (monomials, binomials) BBa6: Simplify and combine like radicals BBa7: Simplify expressions involving radicals by changing to fractional exponents and using exponent laws Negative Exponents BBb: Use reciprocal rule to eliminate denominator negative exponents BBb3: Simplify negative powers of sums and differences BBb4: Evaluate or simplify expressions involving negative exponents Absolute Value BBc1: Evaluate numerical expressions involving absolute values BBc: Recognize the graph of y = x and its variations BBc3: Solve equations of the type ax + b + c = d Linear Inequalities BBd1: Solve or graph linear inequalities in one variable BBe1: Solve consistent x systems by substitution or elimination BBe: Interpret x systems graphically BBe3: Recognize inconsistent or dependent systems algebraically or graphically Verbal Problems BBf1: Work/rate BBf: Distance/rate/time BBf3: Mixture BBf4: Number (reciprocal, digit, consecutive, integer, etc.) BBf5: Area and perimeter BBf6: Other miscellaneous problems BBf5: Area and perimeter
3 Chapter 3 Algebra 3.1 Rational expressions BAa1: Reduce to lowest terms We need to recall the rule for simplifying fractions. To do so, we need to factor out common factors and then we can divide out common factors. Doing it any other way is not supported by the rules of arithmetic. 1.. x(x + ) x(x + 4) The numerator is factored. We need to factor out the denominator. We see that the second factor contains common factor, so we factor it out and divide all common factors x(x + ), assuming x(x + ) 0: = = x(x + ) x(x + ) x(x + ) x(x + ) 1 = 1. x (x + ) x(x + 4) We factor out from the denominator, then we divide out all common factors on the second step. We can do it only if x 0 = x (x + ) x(x + ) = x(x + ) x(x + ) 1 x = 1 x. 3
4 3.1. RATIONAL EXPRESSIONS CHAPTER 3. ALGEBRA 3. (x + 9)(x ) x(3x + 7) We factor 3 from the denominator. Then we divide out all common factors, shown on separate fractions and divide out them (assuming x 9 0). = (x + 9)(x ) x[3(x + 9)] = x + 9 x + 9 x 3x = x 3x. a(3x ) (x + 3)(6 9x) We observe if we reverse the signs in the parenthesis and factor out 3, we may have common factors to divide out. We do so and have: = a(3x ) (x + 3)( 3)(3x ) We divide out (3x ) assuming that (3x ) 0. a = 3(x + 3). x(x + x + 1) x(x + ) We can factor the numerator further seeing that x + x + 1 = (x + 1). There is only one factor we can factor out from the denominator  the number. On the second step, we divide out all common factors. = x(x + 1) x(x + 1) We assume x(x + 1) 0. = x(x + 1) x(x + 1) 1 (x + 1) = 1 (x + 1). 4
5 CHAPTER 3. ALGEBRA 3.1. RATIONAL EXPRESSIONS 3.1. BAa: Add, subtract, multiply, and divide To add or subtract rational expression, we need to find the Least Common Denominator. We multiply all expressions by the appropriate factors and then we add or subtract the resulting numerators. 1. (x + ) x(x + 4) + x The LCD is x(x + 4). = ( (x + ) x(x + 4) ) ( ) x x(x + 4) + x(x + 4) = (x + ) + x[x(x + 4)] x(x + 4) = x3 + 4x + x + 4. x(x + 4). (x + 3) x(x + 4) x = LCD = x(x + 4) = (x + 3) x[x(x + 4)] x(x + 4) x(x + 4) = x + 6 x3 4x x(x + 4) = x3 4x + x + 6. x(x + 4) 3. (x + 3) x(x + 4) x + x = (x + 3)x x(x + 4)(x + ) We divide out a common factor x, assuming (x 0). = x + 3 (x + 4)(x + ). 4. (x + 3) x(x + 4) x + x = = (x + 3) ( + x) x(x + 4) x = (x + 3)(x + ). x (x + 4) = 5
6 3.1. RATIONAL EXPRESSIONS CHAPTER 3. ALGEBRA 5. (x 3) x(x + 4) x + x LCD = x(x + 4)( + x) = (x 3)( + x) x(x + 4)( + x) x([x(x + 4)] x(x + 4)( + x) = x x 6 x 4 4x x(x + )(x + 4) = x4 3x x 6 x(x + )(x + 4) = x4 + 3x + x + 6 x(x + )(x + 4). Note we put the minus sign in front of the fraction and all minuses were replaced by pluses. 6
7 CHAPTER 3. ALGEBRA 3.1. RATIONAL EXPRESSIONS BAa3: Find and simplify products of the form (ax + b)(c/x + d). We need to multiply expressions containing two terms. A convenient way of remembering how to do it is using FOIL method. We will show it in detail for # 1. ( ) 1. (3x + ) x + 5 = ( 3x ) ( + (3x 5) + ) + ( 5) x x = x + 4 x + 10 ( ) 1. (x + ) x ( ) 4 3. (5x 7) 3x + 5 ( ) 7 4. ( x + ) x 5 ( ) 5. (7x ) x + 15 = 0 3 = 15x x. = 1 x + x 4 = x 3 + x. + 5x 8 3x 35 = 5x x. = 7 + 5x + 14 x 10 = 5x x. = x + 4 x 30 = 105x x. 7
8 3.1. RATIONAL EXPRESSIONS CHAPTER 3. ALGEBRA BAa4: Evaluate 1. If x =, then ( ( x) x 5 ) = (+) (+) 5 = 0 3 = 0.. If x =, then 3. If x = 5, then ( ) ( x) x 5 ( ) ( x) x 5 ( ) (x x) 4. If x = 7, then x 5 = ( ) ( ) 5 = 4 7 = 4 7. = (+5) (+5) 5 = 3 0 = If x = 3, then ( ) ( x ) x 5x = ( 7) ( 7) ( 7) 5 = ( 3) ( 3) 5( 3) = = 7 4. = = Note the use of parentheses for negative numbers. This way you will not forget to change the sign if you need to. 8
9 CHAPTER 3. ALGEBRA 3.1. RATIONAL EXPRESSIONS BAa5: Simplify complex fractions We can look at the numerator of the complex fraction as a number, which we try to divide by the denominator fraction. Dividing by a fraction means that we multiply by the reciprocal of that fraction. This step is shown only in # Reduce: ( ) (x+)(x+5) x+3 ( (x+)(x+5) x+4 ) = = (x + )(x + 5) x + 3 (x + )(x + 5) x + 3 = x + 4 x + 3. (x + )(x + 5) x + 4 x + 4 (x + )(x + 5). Reduce: ( ) (x +)(x+5) ( x+3 x+ (x+4)(x+1) = (x + )(x + 5) x + 3 = (x + )(x + 5) x + 3 ) x + (x + 4)(x + 1) (x + 4)(x + 1) x + = (x + )(x + 5)(x + 4)(x + 1). (x + 3)(x + ) 3. Reduce: ( ) x(x+)(x +5) ( x+3 x+ x(x+4) = x(x + )(x + 5) x + 3 ) x + x(x + 4) We factor out the numerator of the numerator fraction. = x(x + )(x + 5) x + 3 = x (x + 5)(x + 4). x x(x + 4) x +
10 3.1. RATIONAL EXPRESSIONS CHAPTER 3. ALGEBRA 4. Reduce: ( ) (x +x)(x+5) x+3 ( x+ x +4x ) = (x + x)(x + 5) x + 3 = x(x + )(x + 5) x + 3 x + x + 4x = x (x + 5)(x + 4). x Reduce: ( ) (x )(x+5) = = = x+3 ( x+ x+4 (x )(x + 5) x + 3 (x )(x + 5) x + 3 ) x(x + 4) x + x + x + 4 x + 4 x + ( x + )(x + 5)(x + 4) (x + 3)( x + ) = (x + 5)(x + 4). x
11 CHAPTER 3. ALGEBRA 3.. FACTORING QUADRATICS 3. Factoring quadratics 3..1 BAb1: Factor a quadratic trinomial in two variables To factor quadratics, one way is to use FOIL method. We look for factors of the format (A 1 a + B 1 b)(a a + B b). Evaluating this, we obtain: (A 1 a + B 1 b)(a a + B b) = A 1 A a + (A 1 B + A B 1 )ab + B 1 B b We organize the information in a table. We will show the process in #1. 1. a + 5ab + 6b Looking at the above expression we can see A 1 A = 1. One natural guess is A 1 = A = 1. We replace A 1 and A in the sum A 1 B + A B 1 = 1B + 1B 1 = B + B 1. The sum should be equal to 5. We finally look at the product B 1 B. It must be equal to 6. On the first step, we factor product B 1 B = 6. The sum is (+5), so we need to look only at positive factors. The possibilities for the factors are (+6, +1) and (+3, +). The first pair of factors has sum (+7) and second one has sum (+5). We see the factors must be (+3) and (+). All the information is organized in the table below. In the left column are shown the two factors of the product in the top row. Their sum is shown in the right column. Product=6 Sum of factors +1, +6 7 NO! +, +3 5 YES! = a + 5ab + 6b = 1 1 a + ( )ab + 3 b = a + ab + 3ab + 6b = a(a + b) + 3b(a + b) = (a + b)(a + 3b).. a 5ab + 6b The only difference between # and # 1 is that the sum B 1 + B = 5, i.e. negative five. It means that the factors are negative. The table shows the values: Product=6 Sum of factors, 3 5 YES! = a ab 3ab + 6b We combine first and second terms, and third and forth terms; then factor: = a(a b) 3b(a b) = (a b)(a 3b). 11
12 3.. FACTORING QUADRATICS CHAPTER 3. ALGEBRA 3. a + 4ab 5b = Product= 5 Sum of factors +1, 5 5 NO! 1, YES! = a + 5ab ab 5b = a(a + 5b) b(a + 5b) = (a + 5b)(a b). 4. 6a + 5ab + b We rearrange this in the following fashion We can now to use the table from #1 = b + 5ab + 6a Product=6 Sum of factors +1, +6 7 NO! +, +3 5 YES! 5. a + 1ab + 0b = b + ab + 3ab + 6a = b(b + a) + 3a(b + 3a) = (b + a)(b + 3a). Product=0 Sum of factors 1, 0 1 NO!, 10 1 YES! = a + ab + 10ab + 0b = a(a + b) + 10b(a + b) = (a + b)(a + 10b). 1
13 CHAPTER 3. ALGEBRA 3.. FACTORING QUADRATICS 3.. BAb3: Factor a perfect square trinomial in more than one variable We need to recognize the perfect square formula: and A + AB + B = (A + B) A AB + B = (A B) = (B A) The last identity resembles the property that the opposite numbers have one and the same square. Example: ( 3) = 9 = 3 1. a 6ab + 9b We see there are differences of numbers squared: A = a, B = 9b = (3b), AB = 6ab = a(3b). Then: = a a(3b) + (3b) = (a 3b). We can check the proper factoring by remultiplying: (a 3b) = a a (3b) + ( 3b) = a 6ab + 9b.. 4a 1ab + 9b A = (a), B = 9b = (3b), AB = 1ab = (a)(3b). Then: = (a) (a)(3b) + (3b) 3. 5a + 0ab + 4b = = (a 3b). A = (5a), B = 4b = (b), AB = 0ab = (5a)(b). Then: = (5a) + (5a)(b) + (b) = (5a + b). 13
14 3.. FACTORING QUADRATICS CHAPTER 3. ALGEBRA 4. a ab 1 + b A = a, B = b = (b 1 ), AB = ab 1. Then: = a ab 1 + b = (a b 1 ). This is a very important example. We see the square here is b = (b 1 ) and a. The equivalent way you could see this is a a/b + 1/b. We can say: a ab 1 + b = a a b + 1 b. Then: (a b 1 ) = (a 1 b ) a + 36ab + 9b A = 36a = (6a), B = 9b = (3b), AB = 36ab = (6a)(3b). Then: = (6a) + (6a)(3b) + (3b) = (6a + 3b). Note! Parentheses are used in all the expressions above. This is first, so we do not get confused, and second, to show not only the variable but also the whole expression in parenthesis is squared. In the cross product, it shows the separated factors to you. In all cases, if you feel you could be confused by notation, use parenthesis for indicating what arithmetic operation you need to do first. 14
15 CHAPTER 3. ALGEBRA 3.. FACTORING QUADRATICS 3..3 BAb4: Recognize relationships among coefficients of a factorable trinomial Find a value of b so that the quadratic factors with integer coefficients, and give the factorization. There may be more than one correct answer. You only need give one correct answer. I will give you all possible answers. For # and #4, there are infinitely many of them. The way these could be computed will be shown in a note at the end of the section for the interested reader. The other items have a finite number of solutions. 1. x + (5 + b)x + 1. We can use FOIL to find all the answers it this case. We look for whole numbers which have product 1 and sum (5+b). So, we factor 1 and show all groups of two factors we may have: (1, 1), (, 6), (3, 4); we could have negative factoring as well: ( 1, 1), (, 6), ( 3, 4). For each pair of factors, we can compute the sum of the factors, it must be (5 + b). We then will have an equation we can solve for b: 5 + b = 1 + 1, b = 8. In a table, we show all possible values for b. If your answer is not among them, check your work.. x + 7x + b. Product=1 Sum=5 + b b 1, , , 4 71, , , We use FOIL method again. Let us have factors x + 7x + b = (x + u)(x + v) = x + vx + ux + uv = x + (u + v)x + uv. We choose one of the factors to be u = 3. Then, using the fact that the left and the right side are simply two representations of the same trimoial, so u + v = 7 and uv = b. We find v first. v = 7 u = 7 3 = 4. Second, we find b. b = uv = 3 4 = 1, 15
16 3.. FACTORING QUADRATICS CHAPTER 3. ALGEBRA b = 6. One solution is x + 7x + b = (x + 3)(x + 4) if b = x + (7 b)x + 4. We use the same (FOIL) method as in #1. The table below shows the values of b for which the trinomial is factored. Product=4 Sum=7 b b 1, , , , , , , , Example for the first row we have b = ( 18). Then: x + [7 ( 18)]x + 4 = x + 5x + 4 = (x + 1)(x + 4). 4. x + x + 9b. We applay the same method as in #: x + x + 9b = (x + u)(x + v) = x + vx + ux + uv = x + (u + v)x + uv. We can choose u =. Then, v = 1. We find then that uv = 9b, or b = uv 9 = ( 1) 9 There is no restriction on b to be integer. We check: = 9. x + x = x + x = (x + )(x 1). Note, that even b is not an integer, the quadratic trinomial still have integer coefficients. 5. x + (1 b)x + 15 In the table is shown the solution using the same idea. 16
17 CHAPTER 3. ALGEBRA 3.. FACTORING QUADRATICS Check (using the last row): Product=15 Sum=1 b b 1, , , , x + (1 9)x + 15 = x 8x + 15 = (x 3)(x 5). Finding the solutions for #: we can choose u N, then from u + v = 7 we find v = 7 u and using FOIL, we compute: u(7 u) = b, We have: b = x + 7x + u(7 u). u(7 u) = x + [u + (7 u)]x + u(7 u) = (x + u)[x + (7 u)]. 17
18 3.3. OPERATIONS ON POLYNOMIALS CHAPTER 3. ALGEBRA 3.3 Operations on Polynomials BAe1: Simplify polynomial expressions involving products of binomials in one or more variables 1. (x 1)(x + ) + x(x 3) We multiply the two binomials using FOIL and expand the second product. On the next step, we combine all the like terms.. (x 1)(x + ) + x(x 3) = x + x x + x 3x = (1 + 1)x + ( 1 3)x = x x. = x + x x + x 3x = (1 + 1)x + ( 1 3)x = x x. 3. (x 1)(x + ) + x(x 3) This is again FOIL method. The only difference is one of the binomials is quadratic. We will not have like terms expanding this product. However, when expanding second product, we may find like terms: = x 3 + x x + x 3 3x = (1 + 1)x 3 x + ( 3)x = x 3 x x. 4. (x )(x + ) x(x 3) You may notice we have a product of sum and difference. There is a formula which you d like to remember: (A + B)(A B) = A B. Here, A = x; B =. Check out the FOIL method used below and see that the crossproducts subtract to (x 1)(x + 7) x(x 3) = x + x x 4 x + 3x = 3x 4. = x + 14x x 7 x + 3x = ( 1)x + ( )x 7 = x + 16x 7. 18
19 CHAPTER 3. ALGEBRA 3.3. OPERATIONS ON POLYNOMIALS 3.3. BAe: Factor a difference of squares We will use the formula A B = (A B)(A+B). I will give you the factors in each case. Note, that in some cases, we can factor out a common factor which is not a binomial. 1. 5x 81y A = 5x = (5x), B = 81y = (9y). We use parenthesis to change the order of arithmetic operations, first we do multiplication, and, then raise to a power. = (5x) (9y) = (5x 9y)(5x + 9y).. 4x 9y A = 4x = (x), B = 9y = (3y) x 144y A = 16x = 4 (x), B = 144y = 16(3y). 4. 5x 36y A = 5x = (5x), B = 36y = (6y). = (x) (3y) = (x 3y)(x + 3y). = 16x 16(3y) = 16[x (3y) ] = 16(x 3y)(x + 3y). = (5x) (6y) = (5x 6y)(5x + 6y). 5. y 4 x 16a b 6 A = y 4 x = (y x), B = 16a b 6 = (4ab 3 ). In this example we have higher powers than two. We see the square could be a power of a variable. Here, A = y x, and B = 4ab 3. = (y x) (4ab 3 ) = (y x 4ab 3 )(y x 4ab 3 ). 19
20 3.3. OPERATIONS ON POLYNOMIALS CHAPTER 3. ALGEBRA BAe4: Divide a third degree by a first or second degree to obtain the quotient and remainder The division of polynomials is very similar to long division of numbers. Note when we have missing terms, in either dividend and the divisor, we have to show these one way or another. First, we can write them with coefficient zero or the other way is to leave an empty space in the dividend when writing it. # The division in detail (including the way one can check his answer) is shown #1 and 1. (x 3 3x + 4) (x + x) The division: x x + x)x 3 + 0x 3x + 4 (x 3 + x ) x 3x ( x 4x) x + 4 reminder = x + x + 4 z + x. We check the division by multiplication. We multiply the answer and the dividend and then we add the reminder. If we worked correctly, we will obtain the divisor.. (3x 3 + 3x + 4) (x x + ) The division: answer divisor + reminder = dividend (x )(x + x) + x + 4 = x 3 + x x 4x + x + 4 = x 3 3x + 4. TRUE! 3x + 6 x x + )3x 3 + 3x + 0x + 4 (3x 3 3x + 6x) 6x 6x + 4 (6x 6x + 1) 8 reminder = 3x x x +.
21 CHAPTER 3. ALGEBRA 3.3. OPERATIONS ON POLYNOMIALS Check: answer divisor + reminder = dividend (3x + 6)(x x + ) 8 = 3x 3 3x + 6x + 6x 6x = 3x 3 + 3x + 4. TRUE! 3. (x 3 + 3x + 4) (x + ) = = x x x (x 3 + 4x 3x + 4) (x 3) = = x + 7x x (x 3 3x + 5x 4) (x ) = = x 3 + 9x 10 x. Once more, here is the way to check the answer. We multiply the quotient and the divisor and add a reminder (example #5): (x )(x 3) + 9x 10 = = x 3 3x 4x x 10 = = x 3 3x + 5x 4. We got the original expression ( the dividend), so that we worked properly. 1
22 3.3. OPERATIONS ON POLYNOMIALS CHAPTER 3. ALGEBRA BAe5: Raise a binomial to the second or third power We can use the following formulas for raising to second and third powers: (A + B) = A + AB + B. Using the formula for the sum, and the fact that B = +( B) we find (A B) = [A + ( B)] = A + A( B) + ( B) = A AB + B. The same way we compute the third power of the sum: Changing B to B we obtain (A + B) 3 = A 3 + 3A B + 3AB + B 3. (A B) 3 = [A + ( B)] 3. = A 3 + 3A ( B) + 3A( B ) + ( B) 3 == A 3 3A B + 3AB B 3. Note that squares are shown with + plus sign regardless whether they are with plus or minus sign in the sum. The only difference when we square a difference is that the crossproduct has negative sign. Note, that in general we do not need two of the formulas: difference squared and difference cubed because these can be derived from the sum squared and the sum cubed by appropriate change of signs. 1. (x + 1) A = x; B = 1. = x + ( x 1) + 1 = x + x + 1. (x + 1) 3 = x x x = x 3 + 3x + 3x (x + ) A = x; B =. = x + x + = x + 4x + 4. (x + ) 3 = x x + 3 x + 3 = x 3 + 6x + 1x + 8.
23 CHAPTER 3. ALGEBRA 3.3. OPERATIONS ON POLYNOMIALS 3. (x 1) A = x; B = 1. = (x) + (x ( 1) + ( 1) = 4x 4x + 1. (x 1) 3 4. (a + b) = (x) (x) ( 1) (x) ( 1) + ( 1) 3 = 8x 3 1x + 6x 1. A = a; B = b. = a + a(b) + (b) = a + 4ab + 4b. (a + b) 3 = a 3 + 3a (b) + 3a(b) + (b) 3 = a 3 + 6a b + 1ab + 8b (x 3) A = x; B = 3. = x + x ( 3) + ( 3) = x 6x + 9. (x 3) 3 = x x ( 3) + 3 x ( 3) + ( 3) 3 = x 3 9x + 7x 7. 3
24 3.4. GRAPHS CHAPTER 3. ALGEBRA 3.4 Graphs BAc: Find the slope of a line from an equation The slope of a line is a fraction slope of a line = coeff x coeff y. Example: for the equation 3y + 4x = 1 we subtract 4x from both sides. 3y + 4x 4x = 1 4x, 3y = 4x + 1. The slope is We can divide by 3: The slope is slope of a line = coeff x coeff y = 4 3. y = 4 3 x = slope = ( 4 3 )/1. 1. The slope of the line y = 5x + 7 is 5. y = 5 1 x The slope of the line 7y = x + 3 is /7 We divide both sides of the equation by 7: y = 7 x The slope of the line 9y 4x + 7 = 0 is 4/9. We subtract from both sides ( 4x + 7). Next, we divide the equation by 9. 9y = 4x 7, y = 4 9 x
25 CHAPTER 3. ALGEBRA 3.4. GRAPHS 4. The slope of the line 3(y 4) = 5(x + 7) is 5/3. We divide by 3. Then, we add 4 to both sides and open the parenthesis. y = 5 (x + 7) + 4, 3 y = 5 3 x , 5 3 x The slope of the line 1y + 4x = 7 is 1/3. We solve for y: 1y = 4x + 7, y = 4 1 x + 7 1, y = 1 3 x
26 3.4. GRAPHS CHAPTER 3. ALGEBRA 3.4. BAc3: Find the slope of a line from the coordinates of two points. We usually denote the slope with m. If we have to denote more than one slope, we could use a subscript (example: m 1 ). We use the formula slope = m = y R y Q x R x Q. where y Q means the y coordinate of Q and all other accordingly. In the formula, we can swap the place of the points P and Q (check!). 1. The coordinates of Q are ( 3, 4) and the coordinates of R are (4, 7). The slope of the line that passes through Q and R is m = y R y Q x R x Q, m = ( 3), m = The coordinates of Q are ( 3, 4) and the coordinates of R are (4, 6). The slope of the line that passes through Q and R is m = ( 3), m = The coordinates of R are ( 3, 4/3) and the coordinates of S are (4, /3). The slope of the line that passes through R and S is m RS = /3 4/3 4 ( 3), m RS = /3 7 = 3 1 7, m RS = 1. 6
27 CHAPTER 3. ALGEBRA 3.4. GRAPHS 4. The coordinates of S are (5/7, 4) and the coordinates of T are (4, 6). The slope of the line that passes through S and T is m ST = /7 = ( 10) 3 7, m ST = ( 10) 3 7, m ST = The coordinates of T are ( 3, 4) and the coordinates of U are (4, 4). The slope of the line that passes through T and U is m TU = ( 4) ( 4), 4 ( 3) m TU = 0. We see that if the y coordinates are the same, the slope is zero. Note if the x coordinates of the two points are the same, then the slope is undefined! This is because division by zero cannot be defined. 7
28 3.4. GRAPHS CHAPTER 3. ALGEBRA BAc4: Find the equation of a line from the coordinates of two points. Given the coordinates of two points, write an equation for the line in both pointslope and slope intercept form if the line has a slope. If the line does not have a slope, write an equation in the form y = C. We have a two phase process. First, we find the slope of the line. We can use the formulas from the previous section. Second, we write pointslope form of the equation. It is y y 0 = m(x x 0 ). We denote here y 0 and x 0 corresponding coordinates of either of the given points. To find interceptslope form, we solve for y. We could open the parenthesis and the final form is y = mx mx 0 + y 0 = mx + (y 0 mx 0 ). 1. The coordinates of Q are ( 3, 4) and the coordinates of R are (4, 7). Equations of the line through Q and R are: For pointslope form, we first compute the slope: m = ( 3) = 3 7. Now, we use the formula with point Q ( 3, 4): y 4 = m[x ( 3)], y 4 = 3 (x + 3). 7 For finding interceptslope form, we solve for y: y = 3 ( 7 x ) + 4, 7 y = 3 7 x
29 CHAPTER 3. ALGEBRA 3.4. GRAPHS. The coordinates of Q are ( 3, 4) and the coordinates of R are (4, 6). Equations of the line through Q and R are: m = ( 3) = 7. We use point R (4,6) for point slope form: y 6 = [x ( 3)], 7 y 6 = (x + 3). 7 The interceptslope form is: y = 7 x , y = 7 x The coordinates of R are ( 3, 4/3) and the coordinates of S are (4, /3). Equations of the line through R and S are: m = /3 4/3 =. 4 ( 3) 1 The point slope form with S (4, /3): The interceptslope form: y 3 = (x 4). 1 y = 1 x , y = 1 x The coordinates of T are ( 3, 4) and the coordinates of U are ( 3, 4). The first coordinates are equal,therefore the slope is undefined. The line has equation x = The coordinates of S are (5/7, 4) and the coordinates of T are (4, 6). The equations of the line through S and T are: The point slope form with S (4, /3) m = /7 = y ( 6) = 70 (x 4). 3 9
30 3.4. GRAPHS CHAPTER 3. ALGEBRA The interceptslope form: y = 70 3 x , y = 70 3 x
31 CHAPTER 3. ALGEBRA 3.4. GRAPHS BAc5: Find the equation of a line given 1 point and the slope. Given the slope of a line and the coordinates of a point on the line, write an equation for the line in both pointslope and slope intercept form. Recall the formulas for the pointslope form and the interceptslope form: y y 0 = m(x x 0 ), y = mx mx 0 + y 0. In the preceding, (x 0,y 0 ) are the coordinates of the given point, the first equation is pointslope form, the second is the interceptslope form. 1. A line passes through (3, 4) and has slope 3. The pointslope equation is y ( 4) = 3(x 3). The interceptslope form of the equation is y = 3x + 9 4, y = 3x A line passes through (7, 4) and has slope 3/7. The pointslope equation is y ( 4) = 3 (x 7). 7 The interceptslope form of the equation is y = 3 7 x + 3 4, y = 3 7 x A line passes through (6, 11) and has slope 1/3. The pointslope equation is y ( 11) = 1 (x 6). 3 The interceptslope form of the equation is y = 1 3 x + 11, y = 1 3 x 9. 31
32 3.4. GRAPHS CHAPTER 3. ALGEBRA 4. A line passes through (/5, 6) and has slope 5. The pointslope equation is y 6 = 5(x 5 ). The interceptslope form of the equation is y 6 = 5x, y = 5x A line passes through (3, 4/9) and has slope 5/9. The pointslope equation is y 4 9 = 5 (x 3). 9 The interceptslope form of the equation is y = 5 9 x , y = 5 9 x
33 CHAPTER 3. ALGEBRA 3.4. GRAPHS BAc6: Determine and/or plot the intercepts of a line, given an equation. Given an equation for a line, give the horizontal and vertical intercepts of the line. To find the intercept means we need to find points such that they are on the line, and one of them has first coordinate zero (0,y 1 ) and the other one has second coordinate zero(x 1, 0). In practice, we replace y with zero and solve for x  xintercept and x with zero and solve for y this is yintercept. yintercept We set x=0 and solve for y ax+by+c=0 (0,y ) 1 (x,0) 1 (0,0) xintercept We set y=0 and solve for x The other way is to plot the line, and then determine the intercepts. For this, we choose any two x coordinates, plug then into the equation of the line and solve for y. We plot the points, and find the horizontal and vertical intercepts. 1. An equation for a line is 3x + 4y = 1. If y = 0, then The horizontal intercept is ( 4, 0). 3x = 1, x = 1 3 = 4. If x = 0, then y = 1, y = 3. The vertical intercept is (0, 3). 33
34 3.4. GRAPHS CHAPTER 3. ALGEBRA. An equation for a line is 5x + 4y = 0. If y = 0, then 5x = 0, x = 4. The horizontal intercept is (4, 0). If x = 0, then y = 0, y = 5. The vertical intercept is (0, 5). 3. An equation for a line is 3x + 4y = 5. If y = 0, then 3x = 5, The horizontal intercept is ( 5/3, 0). x = 5 3 = 5 3. If x = 0, then The vertical intercept is (0, 5/4) y = 5, y = An equation for a line is 3(x ) + 4(y 7) = 1. If y = 0 then 3(x ) + 4(0 7) = 1, 3x = 1, 3x = 1, 3x = 34 x = The horizontal intercept is ( 34/3, 0). 34
35 CHAPTER 3. ALGEBRA 3.4. GRAPHS If x = 0, then 3(0 ) + 4(y 7) = 1, 6 + 4y 8 = 1, 4y = 1, The vertical intercept is (0, 17/). 4y = 34, y = 34 4 = An equation for a line is x + 4y = 13. If y = 0, then x = 13, x = 13. The horizontal intercept is (13, 0). If x = 0, then 0 + 4y = 13, y = The vertical intercept is (0, 13/4). 35
36 3.4. GRAPHS CHAPTER 3. ALGEBRA BAc9: Given an equation determine if a point is above or below the graph. Given an equation for a curve in the Cartesian plane and the coordinates of a point in the plane, determine if the point is above or below the curve. Example: An equation for a line is 3x+4y = 1. The point with coordinates (4, 5) is above this line because the point (4, 0) is on the line and 0 < 5. In general, we evaluate the function at the point x. We compare the value of the function to the y coordinate of the point. Geometrically, we plot the function and a vertical line at horizontal coordinate x 0. If computed value of the function is bigger than the second coordinate of the point, then the point is below the graph. I will denote <>= the process of comparing of two numbers. The expression y(1) means that I evaluate the function y at the point when x = Is the point (0, 4) above, below, or on the curve y = x 3 + x 1? We compute the value of y at x = 0, and compare it with second coordinate of the point specified  4. Is y(0) = = 1 <>= 4? y(0) = 1 < 4. We see that y(0) is less than 4. It means that the point (0, 4) is above the graph.. Is the point ( 3, 1) above, below, or on the curve y = x + x 1? Is y( 3) = ( 3) + ( 3) 1 = <>= 1? y( 3) < 1. The point ( 3, 1) is above the graph y = x + x 1 because y( 3) = 5 < Is the point (1, 1) above, below, or on the curve y = x 3 + x? Is y(1) = <>= 0? y(1) = 1. The point (1, 1) is on the graph y = x 3 + x because y(1) = Is the point (3, 4) above, below, or on the curve y = x 3 + x 100? Is y(3) = <>= 4? y(3) < 4. The point (3, 4) is above the graph because y(3) = 67 < 4. 36
37 CHAPTER 3. ALGEBRA 3.4. GRAPHS 5. Is the point (16, 50) above, below, or on the curve y = 5x + x 1? Is y(16) = = <>= 50? y(16) > 50. The point (16, 50) is below the graph y = 5x + x 1 because y(16) = 87 >
38 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA 3.5 Quadratic/Rational Equations BAd1: Quadratics of the form ax + b = 0 or (ax + b) + c = 0. Solve a quadratic equation where the square has been completed. It is much easier to solve a quadratic equation once the square has been completed. We simply move the free term on the other side and take square root from both sides of the equation. We have two equations with one and the same left side and opposite right sides. 1. Solve 4x 36 = 0 for x. We add the free term 36 to both sides of the equation: 4x = We solve for x: x = 36 4 = 9, x = 9 = 3. Recall that when taking square root we compute the principal square root, i.e. the square root without its sign. There are two numbers with absolute value of 3: x 1 = 3 and x = 3. The solutions are: x 1 = +3 and x = 3.. Solve 4(x 1) 36 = 0 for x. We solve these equations and find The solutions are x 1 = 15 and x = 9. 4(x 1) = 36, (x 1) = 36 4 = 9, x 1 = + 9 = +3 or x 1 = 9 = 3. x 1 = 3, x 1 = 15. x 1 = 3, x = Solve 3a + 36 = 0 for a. 3a = 36, a = 36 3 = 1, 38
39 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS a 1 = + 1 = + 3, a = 1 = 3. The solutions are a 1 = + 3 and a = 3. Note, that the solutions may not be rational numbers so that the FOIL device is not always useful. 4. Solve 5(x + 3) 5 = 0 for x. 5(x + 3) = 5, (x + 3) = 5 5 = 5, x + 3 = + 5 or x + 3 = 5. x 1 = 3 + 5, x = 3 5 The solutions are x 1 = and x = Solve (x + 9) 16 = 0 for x. The solutions are x 1 = 0 and x = 18. (x + 9) = 16, (x + 9) = 16 = 81, x + 9 = + 81 = +9 or x + 9 = 81 = 9. x 1 = = 0, x = 9 9 =
40 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA 3.5. BAd: Quadratics of the form ax + bx + c = 0 with a 1 by factoring over the integers. Solve a quadratic equation where the quadratic expression factors over the integers. To solve quadratic equation we use zero product principle: the product of two numbers is zero if and only if at least one of the factors is zero. For quadratics, to find the factors of the quadratic expression and apply the principle means to make each factor equal to 0. If we have rational solutions to the equation, we can use FOIL method for factoring. The details will be shown with #1. Be advised this is a method of trials, so you must be consistent and try until you find the answer or exhaust all possibilities. Check your work carefully to avoid making errors. 1. Solve x + 5x 3 = 0 for x. FOIL method for factoring. We need two First terms with product (+): the only combination is (+, +1). We need two Last terms with product ( 3): they could be (+1, 3) and ( 1, +3). We compute the sum Outer terms product plus Inner terms product until we have a sum equal to (+5). In the table, FTPr means First terms product, LTPr means Last terms product and OIpSum means Outer and Inner terms product sum. Outer terms are shown in outer positions, and Inner terms in inner. The information is organized in the following table. For example the first row represents the quadratic in factors product [(+1)x + (+1)] [(+)x + ( 3)]  the terms come in the order they are in FOIL. FTPr= LTPr=3 OIpSum=+5 +1, + +1, , + 1, , +1 +1, , +11, We see we have the sum we need in the last row. Then, the equation factors 0 = x + 5x 3 = (x 1)(x + 3). We now can solve x 1 = 0 and x + 3 = 0. The first one gives the solution x = 1 and second x = ( 3). For all the following problems, only the table, resulting equations and solutions will be shown.. Solve 4x + 4x 3 = 0 for x. 40
41 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS The equivalent forms of the equation are The solutions are x = 1 and x = Solve 6a + 13a + 6 = 0 for a. The equivalent forms of the equation are FTPr=4 LTPr=3 OIpSum=+4 +1, +4 +1, , +41, , +1 +1, , +11, , + +1, , + 1, = 4x + 4x 3 = (x 1)(x + 3). FTPr=6 LTPr=6 OIpSum=+13 +1, +6 +, , +6 +3, , +6 +6, , +6 +1, , +3 +1, , +3 +3, = 6a + 13a + 6 = (a + 3)(3a + ). The solutions of the equation are a 1 = 3 and a = Solve 6x x 15 = 0 for x. The equation is FTPr=6 LTPr=15 OIpSum=1 +1, +65, , +6 +3, , +35, , +3 +3, = 6x x 15 = (x + 3)(3x 5). The solutions of the equation are x 1 = 3 and x =
42 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA 5. Solve 6x + 13x 15 = 0 for x. We can use the table above to solve # 5 as well. The table is reduced to only one row: The equation is FTPr=6 LTPr=15 OIpSum=+13 +1, +6 +3, = 6x + 13x 15 = (x + 3)(6x 5). The solutions of the equation are x 1 = 3 and x =
43 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS BAd3: Quadratics using the quadratic formula, where the discriminant might be positive, negative or zero. Solve a quadratic equation where the roots may not be rational numbers. Either we complete the square or use the quadratic formula. We use the quadratic formula. It can be used in every case. If the equation is ax + bx + c = 0 and we solve for x, the solutions of the equation are This means there are two solutions: and x 1, = b ± b 4ac. a x 1 = b + b 4ac a x = b b 4ac. a The other general way for solving quadratic equations is by completing the square. We ilustrate it in the following problem. 1. x + 5x 4 = 0. We add 4 to both sides of the equation. x + 5x = 4. We divide both sides by. x + 5 x =. Here is why this technique is called completing the square. On the left side we want to have a perfect square. The first term is x. We know the crossproduct is 5 x. This is the double product of the first and the second term. We know that the first one is x and determine the second one (called below b): x b = 5 x. We solve for b: b = 5. Therefore we need to add to both sides the square of b,i.e.  4 ( 5 4 ) : x x + ( 5 4) = + 43 ( ) 5. 4
44 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA We factor the square on the left. ( x + 5 ) = We take square roots from both sides and have: or x = x = We have plus or minus because there are two numbers with square 57/16, just as (+3) = ( 3) = 9. We solve for x: x 1 = , x = x 4x 5 = 0. We will use the same method for this equation. or 4x 4x = 5, (x) (x) = 5 + 1, (x + 1) = 6. x + 1 = + 6 x + 1 = 6. The solutions are x 1 = ( 1 + 6)/ and x = ( 1 6)/. 3. 6a + 13a + 7 = 0: Here we will use the quadratic formula. Since a = 6, b = 13 and c = 7 we use the formulas: x 1 = b + b 4ac a and x = b b 4ac. a x 1 = (+13) + (+13) 4 (+6) (+7), 6 44
45 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS x 1 = 13 ± , 1 x 1 = = 1. 1 For x we have x = 13 1 = Then, the solutions are x 1 = 1 and x = 7/6. 4. x + x + 15 = 0. x 1 = , 1 x 1 = Recall the fact that 1 = i is an imaginary number which we assume has square equal to ( 1). We can write then 56 = ( 1)56 = 1 56 = 56i. We can factor out : so the solution is The same way + 56i x 1 = + 56i. = ( i = 14i x 1 = = i. x = 1 14i. The solutions are x 1 = 1 + i 14 and x = 1 i x + 13x + 15 = 0. The same way: x 1 = , 6 x 1 = , 1 x 1 = 13 + i x = 13 i The solutions are x 1 = ( 13 + i 191)/1 and x = ( 13 i 191)/1. 45
46 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA BAd4: Equations quadratic in u, where u = x, x 3,... Solve an equation which is quadratic in a power of x. To solve an equation which is quadratic in a power of x, we make use of the property of exponents as in example shown below: (x ) 3 = (x 3 ) = x 3 = x 6. The property states if we have a power of a variable (say x), we can break the exponent down into two factors and when computing this way, the result is the same. You can check for yourself ( ) 3 = ( 3 ) = 6. We can use this rule for rational exponents as well. Recall the denominator of the exponent is the root which we take to compute the number. 1. x /3 + 3x 1/3 + = 0. We see that = 1. We can simplify the equation by letting 3 3 This is our substitution. Then, we have The equation becomes x 1/3 = u. u = x /3. u + 3u + = 0. We see we can solve this equation easily (u+1)(u+) = 0. The solutions are u 1 = 1 and u =. The equation is not solved yet! We need to go back and use the substitution to find x: We find The other solution u = will give us x 1/3 = 1. x 1 = 1. x 1 3 =, x = 8.. x 6 + 3x 3 + = 0. We substitute The equation becomes x 3 = v. x 6 + 3x 3 + = (x 3 ) + 3(x 3 ) + = v + 3v + = 0. 46
47 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS The last right equation is solved above. We have the solutions v 1 = 1 and v =. We now solve x 3 = 1, x = 0. We factor this equation and find (x + 1)(x x + 1) = 0. Now, we use zero product principle and find x + 1 = 0, Next, we solve x 1 = 1. x x + 1 = 0. Using the quadratic formula, we find two complex number roots: The other root is x = 1 + ( 1) 4 1 1, x = 1 + i 3. x 3 = 1 i 3. For the second solution of u we solve the same way: This equation has three other roots: x 3 =, x 3 + = 0, (x + 3 )(x 3 x + ( 3 ) ) = 0. x 4 = 3. To ease the computations, we first compute the discriminant of (x 3 x+( 3 ) ) = 0: D = ( 3 ) 4 1 ( 3 ) ) = 3 3 ( ). The remaining solutions are ( ) x 5 =, 3 x 5 = + i 3 3 and 3 x 6 = i 3 3. All the solutions of the equation x 6 + 3x 3 + = 0 are x 1 = 1, x = 1/ + i 3/, x 3 = 1/ i 3/, x 4 = 3, x 5 = 3 (1/ + i 3/) and x 6 = 3 (1/ i 3/). 47
48 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA 3. x 4 3x + 1 = 0. We substitute x = u. u 3u + 1 = 0. We use FOIL and find (u 1)(u 1) = 0. Thus, u 1 = 1 and u = 1. We find all the solutions for x now: x = 1, 1 x 1, = ± = ±. The first two solutions are x 1 = / and x = /. Finally, we solve x = 1, x 3,4 = ±1. The other two solutions are x 3 = 1 and x 4 = x x 1 = 0. This equation is not in a quadratic form for any power of x. But, we can transform it in this form by multiplying it by x: (x x 1 )x = 0 x x + 5x + 4 = 0 This is a regular quadratic equation and we solve it for x: The solutions are x 1 = 4 and x = 1. (x + 4)(x + 1) = 0 5. x + x = 0. We use the same idea for this equation (but multiplying by x ): We substitute x = u: x 4 x + 1 = 0 u u + 1 = 0 This equation is (u 1) = 0. Therefore, there is only one repeated solution u 1, = 1. We solve for x now and find x 1 = 1 and x = 1. The original equation is not a polynomial equation so that it has only two solutions as stated above. 48
49 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS BAd5: Rational equations leading to quadratic equations. Solve a rational equation which can be reduced to a quadratic equation. We need to transform the rational equation to quadratic equation by using a process known as clearing fractions. We multiply the equation by the Least Common Denominator of all rational fractions in the equation. This is not all. We need to list all restrictions, i.e., all values of x for which the equation is defined. Recall we cannot divide by zero. 1. x 5 x + 1 = x +. We see that the LCD = x + 1. This denominator must not be equal to zero, so we need x 1. We multiply now the equation by (x + 1): We simplify it and solve for x: x 5 = (x + )(x + 1). x 5 = x + 3x +, We use the quadratic formula: x + x + 7 = 0. and x 1 = There are two complex number solutions: x = and x 5. x + 1 = x + x 1. LCD = (x + 1)(x 1). We restrict x: x 1 and x +1. We can clear the fractions now: x 1 = 1 + i3 3 x = 1 i3 3. (x 5)(x 1) = (x + )(x + 1), 49
50 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA We use quadratic formula: x 7x + 5 = x + 3x +, x 10x + 3 = 0. and The solutions are and x 5 3. = 3x +. x + 1 LCD = x + 1, then x 1. Clearing fractions: The solutions are and i.e. x 1 = 10 + ( 10) x = 10 ( 10) x 1 = 5 + x = 5. x 5 = (3x + )(x + 1), x 5 = 3x + 5x +, 3x + 3x + 7 = 0. x 1 = x = , 3 4. and x 1 = 1 + i 75 6 x = 1 i x x 1 = 3. LCD = (x + 1)(x 1), thus x 1 and x +1. Clearing fractions: [ 1 x ] (x + 1)(x 1) = 3(x + 1)(x 1). x 1 50
51 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS We open the parenthesis on the left and remove the common factors: We solve for x: The solutions are: and i.e. and 5. Solve 3x 5 = x 1. x 1 LCD = x 1, thus x 1. Clearing fractions: The solutions are: and i.e. and (x 1) + (x + 1) = 3(x 1). 3x 3 = x, 3x x 3 = 0. x 1 = + ( ) 4 3 ( 3) 3 x 1 = ( ) 4 3 ( 3), 3 x 1 = x = x 5 = (x 1)(x 1), 3x 5 = x 3x + 1, x 6x + 6 = 0, x 3x + 3 = 0. x 1 = 3 + ( 3) x = 3 ( 3) 4 1 3, 1 x 1 = 3 + i 3 x = 3 i 3. 51
52 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA BAd6: Rational equations leading to linear equations. Solve a rational equation that reduces to a linear equation. There is no difference between solving rational equations leading to quadratic equations and those leading to linear equations. We simply have only one solution. 1. x + 5 x + 1 = x +. LCD = x + 1, thus x 1/. Clearing fractions: x + 5 = (x + )(x + 1), x + 5 = x + 5x +. We solve for x: 5x = 3, x = 3 5. x 5. x + 1 = x + x 1. LCD = (x + 1)(x 1), thus x 1/ and x 1. Clearing fractions and solving: (x 5)(x 1) = (x + 1)(x + ), x 7x + 5 = x + 5x +, 1x = 3, x = x 5 3. = 3x +. x + 1 LCD = x + 1, thus x 1. Clearing fractions and solving: 3x 5 = (3x + )(x + 1), 3x 5 = 3x + 5x +, 5x = 7, x =
53 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS 4. x x x 1 = 1. LCD = (x + 1)(x 1), thus x ±1. We clear fractions and solve: [ x x ] (x + 1)(x 1) = 1 (x + 1)(x 1), x 1 We have now x(x 1) + 1 (x + 1) = x 1, x x + x + 1 = x = 1 FALSE! We see this equation does not have any solution. 4x 5x = x 1. x 1 LCD = x 1, thus x 1/. Clearing fractions and solving: 4x 5x + 7 = (x 1)(x 1), 4x 5x + 7 = 4x 4x + 1, x = 6, x = 6. 53
54 3.5. QUADRATIC/RATIONAL EQUATIONS CHAPTER 3. ALGEBRA BAd7: Solve literal equations involving rational expressions. Solve a rational equation in several variables for a specific variable. We use the same method when solving literal equations as when we use numbers. The only difference is we keep the expression with all variables. Keep in mind variables represent numbers and, if we cannot do something with numbers, we cannot do it with variables too (example: dividing by zero is not defined so we cannot allow to have denominators equal to zero). Solve each of the following equations for b b + 1 a = 1. We isolate on the first step 1/b by subtracting 1/a from both sides and simplify: 1 b = 1 1 a = a 1 a. We take the reciprocal of both sides (after restricting a 1 since 1/b cannot be zero): b = a a 1. a b + c d = 1. To isolate b on one side, we first subtract the fraction c/d from both sides and simplify: a b = 1 c d = d c d. We multiply both sides by db ( after restricting c d): d c Dividing out common factors will give us a b b d c d = c d d d c d b. b = ad d c b + 1 a = 1 a + b. We can clear fractions as we have it done before, restricting a 0 b 0 and a b. We can multiply by ab(a + b): a(a + b) + b(a + b) = ab, 54
55 CHAPTER 3. ALGEBRA 3.5. QUADRATIC/RATIONAL EQUATIONS a + ab + ab + b = ab. We subtract ab from both sides of the equation and see that we have a quadratic equation for b. We solve it for b using quadratic formula: Then, b + ab + a = 0, b 1, = a ± a 4 1 a. 1 b 1 = a 3 a + i, b = a 3 a i. a b + b a = 1. We clear fractions after restricting a 0 and b 0: a b ab + b a = 1ab, a + b = ab. We subtract ab from both sides of the equation. This is a quadratic equation for b. We solve it using quadratic formula: Then, b ab + a = 0, b 1, = ( a) ± ( a) 4 1 a. 1 b 1 = a 3 a + i, b = a 3 a i. 1 b 1 a =. The restrictions are a 0 and b 0. We can proceed as in #1: we isolate 1/b on one side of the equation, and simplify. 1 b = + 1 a = a + 1. a We cannot have the left side zero, so we restrict the right side not to be equal to zero a 1/. We then take the reciprocal of both sides: b = a a
56 3.6. RADICALS AND FRACTIONIAL EXPONENTS CHAPTER 3. ALGEBRA 3.6 Radicals and Fractionial exponents BBa4: Express radicals in terms of fractional exponents and vice versa. Rewrite each expression using the alternate form for expressing roots. Do not simplify. Recall that A 1/n is actually n A and ( n B) m is B m/n : and A 1/n = n A ( n B) m = B m/n m = 1 and n = m = 1 and n = / m = 3 and n = / n = and m = ( 3) /5 m = and n = 5. = 36 1/. = ( 7) 1/3. = ( 100) 3. = ( 5) 5. = ( 5 3). 56
57 CHAPTER 3. ALGEBRA 3.6. RADICALS AND FRACTIONIAL EXPONENTS 3.6. BBa1: Evaluate numerical expressions involving radicals and/or fractional exponents. Simplify numerical expressions involving roots in either radical or fractional exponent form = = / = ( 100) 3 = 10 3 = / = ( 5) 5 = 5 5 = ( 3) /5 = ( 5 3) = ( ) = 4. 57
58 3.6. RADICALS AND FRACTIONIAL EXPONENTS CHAPTER 3. ALGEBRA BBa: Simplify radical expressions. Reduce and combine radicals in each radical expression. You may assume all variables represent positive quantities. Recall a perfect square is a number or expression which is a square of a rational number or another expression. Example 4 = and a 4 = (a ). The same way we can define perfect third power, fourth power and so on. In general, perfect nth power is a number or expression which is an nth power of a number or expression x 5 y 3 /z 6 We see that 16 = 3 and x 5 = x 3 x. The other expressions y 3 and z 6 are perfect third powers (also known as perfect cube). = 3 ( 3 x 3 y 3 /z 3 )(x ) = 3 (xy/z ) 3 3 x = ( xy z )3/3 3 x = xy z 3 x x 5 y 9 /z 6 = 4 (3 4 x 4 y 8 /z 4 )(xy/z ) = [(3xy ) 4 ] 1/4 4 xy/z = 3xy 4 xy z z = = = x 7 y 3 /z 8 = 3 (3 3 x 6 y 3 /z 6 )(x/z ) = [(3x y/z ) 3 ] 1/3 3 x/z = 3x y 3 x z z. 5. 1x 4 y 3 = ( x 4 y )(3y) = [ (x 4 y) ] 1/ 3y = x y 3y. 58
59 CHAPTER 3. ALGEBRA 3.6. RADICALS AND FRACTIONIAL EXPONENTS BBa3: Rationalize the numerator or denominator of a quotient involving radicals. To rationalize the numerator or denominator of an expression, we use the following formula: (A + B)(A B) = A B. We will multiply by an expression of one, containing the conjugate of the radical number. The conjugate of A + B is A B, and the conjugate of A B is A + B. 1. x y (x y) The conjugate of ( x y) is ( x + y). Therefore, we need to multiply with expression of one x+ y x+ y : x + y. x + y The conjugate is x y. = x y x + y x y x + y = ( x) ( y) (x y)( x + y) = x y x y 1 x + y = = = 1 x + y, if x y. x + y x y x + y x y (x y) (x + y)( x, if x y. y) 3. x + x 4 The conjugate is x + + x. x + x x + + x = 4 x + + x = ( x + ) ( x ) 4( x + + x ) x + x + = 4( x + + x ) 1 =. x + + x 59
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