Example Solve for x in terms of a and b:2x + b = a Subtract b from both sides of the equation: 2x + b – b = a – b Divide both sides of the equation by 2: Solving Inequalities Solving inequalities is very similar to solving equations. The four symbols used when solving inequalities are as follows: ■ Ͻ is less than ■ Ͼ is greater than ■ Յ is less than or equal to ■ Ն is greater than or equal to When solving inequalities, there is one catch: If you are multiplying or dividing each side by a negative number, you must reverse the direction of the inequality symbol. For example, solve the inequality –3x + 6 Յ 18: 1. First subtract 6 from both sides: 2. Then divide both sides by –3: 3. The inequality symbol now changes: Solving Compound Inequalities A compound inequality is a combination of two inequalities. For example, take the compound inequality –3 Ͻ x + 1 Ͻ 4. To solve this, subtract 1 from all parts of the inequality. –3 – 1 Ͻ x + 1 – 1 Ͻ 4 – 1. Simplify. –4 Ͻ x Ͻ 3. Therefore, the solution set is all numbers between –4 and 3, not including –4 and 3. x ՆϪ4 –3x –3 Յ 12 –3 –3x ϩ 6 Ϫ 6 Յ 18 Ϫ 6 x ϭ a Ϫ b 2 2x 2 ϭ a Ϫ b 2 – ALGEBRA– 343 Multiplying and Factoring Polynomials When multiplying by a monomial, use the distributive property to simplify. Examples Multiply each of the following: 1. (6x 3 )(5xy 2 ) = 30x 4 y 2 (Remember that x = x 1 .) 2. 2x (x 2 – 3) = 2x 3 – 6x 3. x 3 (3x 2 + 4x – 2) = 3x 5 + 4x 4 – 2x 3 When multiplying two binomials, use an acronym called FOIL. F Multiply the first terms in each set of parentheses. O Multiply the outer terms in the parentheses. I Multiply the inner terms in the parentheses. L Multiply the last terms in the parentheses. Examples 1. (x – 1)(x + 2) = x 2 + 2x – 1x – 2 = x 2 + x – 2 F O I L 2. (a – b) 2 = (a – b)(a – b) = a 2 – ab – ab – b 2 F O I L Factoring Polynomials Factoring polynomials is the reverse of multiplying them together. Examples Factor the following: 1. 2x 3 + 2 = 2 (x 3 + 1) Take out the common factor of 2. 2. x 2 – 9 = (x + 3)(x – 3) Factor the difference between two perfect squares. 3. 2x 2 + 5x – 3 = (2x – 1)(x + 3) Factor using FOIL backwards. 4. 2x 2 – 50 = 2(x 2 – 25) = 2(x + 5)(x – 5) First take out the common factor and then factor the difference between two squares. Solving Quadratic Equations An equation in the form y = ax 2 + bx + c,where a, b, and c are real numbers, is a quadratic equation. In other words, the greatest exponent on x is two. – ALGEBRA– 344 Quadratic equations can be solved in two ways: factoring, if it is possible for that equation, or using the quadratic formula. By Factoring In order to factor the quadratic equation, it first needs to be in standard form. This form is y = ax 2 +bx + c. In most cases, the factors of the equations involve two numbers whose sum is b and product is c. Examples Solve for x in the following equation: 1. x 2 – 25 = 0 This equation is already in standard form. This equation is a special case; it is the difference between two perfect squares. To factor this, find the square root of both terms. The square root of the first term x 2 is x. The square root of the second term 25 is 5. Then two factors are x – 5 and x + 5. The equation x 2 – 25 = 0 then becomes (x – 5)(x + 5) = 0 Set each factor equal to zero and solve x – 5 = 0 or x + 5 = 0 x = 5 or x = –5 The solution is {5, –5}. 2. x 2 + 6x = –9 This equation needs to be put into standard form by adding 9 to both sides of the equation. x 2 + 6x + 9 = –9 + 9 x 2 + 6x + 9 = 0 The factors of this trinomial will be two numbers whose sum is 6 and whose product is 9. The fac- tors are x + 3 and x + 3 because 3 + 3 = 6 and 3 × 3 = 9. The equation becomes (x + 3)(x + 3) = 0 Set each factor equal to zero and solve x + 3 = 0 or x + 3 = 0 x = –3 or x = –3 Because both factors were the same, this was a perfect square trinomial. The solution is {–3}. 3. x 2 = 12 + x This equation needs to be put into standard form by subtracting 12 and x from both sides of the equation. x 2 – x – 12 = 12 – 12 + x – x x 2 – x – 12 = 0 – ALGEBRA– 345 Since the sum of 3 and –4 is –1, and their product is –12, the equation factors to (x + 3) (x – 4) = 0 Set each factor equal to zero and solve: x + 3 = 0 or x – 4 = 0 x = –3 or x = 4 The solution is {–3, 4}. By Quadratic Formula Solving by using the quadratic formula will work for any quadratic equation, especially those that are not fac- torable. Solve for x: x 2 + 4x = 1 Put the equation in standard form. x 2 + 4x – 1 = 0 Since this equation is not factorable, use the quadratic formula by identifying the value of a, b, and c and then substituting it into the formula. For this particular equation, a = 1, b = 4, and c = –1. The solution is . The following is an example of a word problem incorporating quadratic equations: A rectangular pool has a width of 25 feet and a length of 30 feet. A deck with a uniform width sur- rounds it. If the area of the deck and the pool together is 1,254 square feet, what is the width of the deck? ͭ –2 ϩ 2 5 , –2 – 2 5 ͮ x ϭ –2 ; 2 5 x ϭ –4 2 ; 22 5 2 x ϭ –4 ; 2 20 2 x ϭ –4 ; 2 16 ϩ 4 2 x ϭ –4 ; 2 4 2 Ϫ 41121–12 2112 x ϭ –b ; 2 b 2 –4ac 2a – ALGEBRA– 346 Begin by drawing a picture of the situation. The picture could be similar to the following figure. Since you know the area of the entire figure, write an equation that uses this information. Since we are trying to find the width of the deck, let x = the width of the deck. Therefore, x + x + 25 or 2x + 25 is the width of the entire figure. In the same way, x + x + 30 or 2x + 30 is the length of the entire figure. The area of a rectangle is length × width, so use A = l × w. Substitute into the equation: 1,254 = (2x + 30)(2x + 25) Multiply using FOIL: 1,254 = 4x 2 + 50x + 60x + 750 Combine like terms: 1,254 = 4x 2 + 110x + 750 Subtract 1,254 from both sides: 1,254 – 1,254 = 4x 2 + 110x + 750 – 1,254 0 = 4x 2 + 110x – 504 Divide each term by 2: 0 = 2x 2 +55x – 252 Factor the trinomial: 0 = (2x + 63 )(x – 4) Set each factor equal to 0 and solve 2x + 63 = 0 or x – 4 = 0 2x = –63 x = 4 x = –31.5 Since we are solving for a length, the solution of –31.5 must be rejected. The width of the deck is 4 feet. Rational Expressions and Equations Rational expressions and equations involve fractions. Since dividing by zero is undefined, it is important to know when an expression is undefined. The fraction is undefined when the denominator x – 1 = 0; therefore, x =1. 5 x Ϫ 1 x x x x 25 30 – ALGEBRA– 347 You may be asked to perform various operations on rational expressions. See the following examples. Examples 1. Simplify . 2. Simplify . 3. Multiply . 4. Divide . 5. Add . 6. Subtract . 7. Solve . 8. Solve . Answers 1. 2. 3. 4. 5. 6. 3x ϩ 18 Ϫ x ϩ 2 3x ϭ 2x ϩ 20 3x 1 ϩ 3x xy a1a ϩ 2 2 1a ϩ 1 21a ϩ 22 ϫ 21a ϩ 1 2 a1a Ϫ 3 2 ϭ 2 a Ϫ 3 4x1x ϩ 4 2 2x 2 1x Ϫ 4 21x ϩ 42 ϭ 2 x1x Ϫ 4 2 1x ϩ 3 21x Ϫ 32 31x Ϫ 3 2 ϭ 1x ϩ 3 2 3 x 2 b x 3 b 2 ϭ 1 xb 1 x ϭ 1 4 ϩ 1 6 2 3 x ϩ 1 6 x ϭ 1 4 x ϩ 6 x – x –2 3x 1 xy ϩ 3 y a 2 ϩ 2a a 2 ϩ 3a ϩ 2 Ϭ a 2 –3a 2a ϩ 2 4x x 2 –16 ϫ x ϩ 4 2x 2 x 2 Ϫ 9 3x Ϫ 9 x 2 b x 3 b 2 – ALGEBRA– 348 . . Answers 1. 2. 3. 4. 5. 6. 3x ϩ 18 Ϫ x ϩ 2 3x ϭ 2x ϩ 20 3x 1 ϩ 3x xy a1a ϩ 2 2 1a ϩ 1 21 a ϩ 22 ϫ 21 a ϩ 1 2 a1a Ϫ 3 2 ϭ 2 a Ϫ 3 4x1x ϩ 4 2 2x 2 1x Ϫ 4 21 x ϩ 42 ϭ 2 x1x Ϫ 4 2 1x ϩ 3 21 x Ϫ 32 31x Ϫ 3 2 ϭ 1x. width of the deck? ͭ 2 ϩ 2 5 , 2 – 2 5 ͮ x ϭ 2 ; 2 5 x ϭ –4 2 ; 22 5 2 x ϭ –4 ; 2 20 2 x ϭ –4 ; 2 16 ϩ 4 2 x ϭ –4 ; 2 4 2 Ϫ 41 121 – 12 21 12 x ϭ –b ; 2 b 2 –4ac 2a – ALGEBRA– 346 Begin by drawing. 2 ϭ 1x ϩ 3 2 3 x 2 b x 3 b 2 ϭ 1 xb 1 x ϭ 1 4 ϩ 1 6 2 3 x ϩ 1 6 x ϭ 1 4 x ϩ 6 x – x 2 3x 1 xy ϩ 3 y a 2 ϩ 2a a 2 ϩ 3a ϩ 2 Ϭ a 2 –3a 2a ϩ 2 4x x 2 –16 ϫ x ϩ 4 2x 2 x 2 Ϫ 9 3x Ϫ 9 x 2 b x 3 b 2 – ALGEBRA– 348