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Algebra and Functions Practice Problems: GED Math (page 2)

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Updated on Mar 23, 2011

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  1. d.   Subtract the like terms by subtracting the coefficients of the terms: 9a – 5a = 4a. 4a and 12a2 are not like terms, so they cannot be combined any further: 9a + 12a2 – 5a = 12a2 + 4a.
  2. a.   Multiply the coefficients of the terms in the numerator, and add the exponents of the bases: (3a)(4a) = 12a2. Do the same with the terms in the denominator: 6(6a2) = 36a2. Finally, divide the numerator by the denominator. Divide the coefficients of the terms and subtract the exponents of the bases: .
  3. c.   To find the product of two binomials, multiply the first term of each binomial, the outside terms, the inside terms, and the last terms. Then, add the products: (x – 3)(x + 7) = x2 + 7x – 3x – 21 = x2 + 4x – 21.
  4. d.   To find the product of two binomials, multiply the first term of each binomial, the outside terms, the inside terms, and the last terms. Then, add the products: (x – 6)(x – 6) = x2 – 6x – 6x + 36 = x2 – 12x + 36.
  5. b.   To find the factors of a quadratic, begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic: –3 and 2 multiply to –6 and add to –1. Therefore, the factors of x2x – 6 are (x – 3) and (x + 2).
  6. c.   To solve the equation, subtract 5 from both sides of the equation, then divide both sides by 9: 9a + 5 = –22, 9a + 5 – 5 = –22 – 5, 9a = –27, a = –3.
  7. d.   Begin with the innermost function: find g(–3) by substituting –3 for × in the function g(x): g(–3) = (–3)2 = 9. Then, substitute the result of that function for x in f(x): f(9) = 2(9) – 1 = 18 – 1 = 17.
  8. b.   Begin with the innermost function: find f(–2) by substituting –2 for x in the function f(x): f(–2) = 3(–2) + 2 = –6 + 2 = –4. Then, substitute the result of that function for x in g(x): g(–4) = 2(–4) – 3 = –8 – 3 = –11.
  9. a.   The domain of the function is all real numbers; any real number can be substituted for x. There are no x values that can make the function undefined. The range of a function is the set of all possible outputs of the function. Because the x term is squared, then made negative, the largest value that this term can equal is 0 (when x = 0). Every other x value will result in a negative value for f(x). The range of f(x) is all real numbers less than or equal to 0.
  10. c.   The square root of a negative value is imaginary, so the value of 4x – 1 must be greater than or equal to 0. 4x – 1 ≥ 0, 4x ≥ 1, x. The domain of f(x) is all real numbers greater than or equal to . Because x must be greater than or equal to , the smallest value of f(x) is the square root of 0, which is 0. The range of the function is all real numbers greater than or equal to 0.
  1. d.   The terms 5a and 7b have unlike bases; they cannot be combined any further. Add the terms in the denominator: b + 2b = 3b. Divide the b term in the numerator by the 3b in the denominator: middle.
  2. b.   Multiply 2x2 and 4y2 by multiplying the coefficients of the terms: (2x2)(4y2) = 8x2y2. 8x2y2 and 6x2y2 have like bases, so they can be added. Add the coefficients: 8x2y2 + 6x2y2 = 14x2y2.
  3. a.   First, multiply (x + 2) by 4: 4(x + 2) = 4x + 8. Then, subtract 3x from both sides of the inequality and also subtract 8 from both sides of the inequality:
      3x – 6 < 4x + 8
      3x – 6 – 3x < 4x + 8 – 3x
      –6 < x + 8
      –6 – 8 < x + 8 – 8
      x > –14
  4. a.   To find the product of two binomials, multiply the first term of each binomial, the outside terms, the inside terms, and the last terms. Then, add the products. (x – 1)(x + 1) = x2 + xx – 1 = x2 – 1.
  5. e.   (x + c)2 = (x + c)(x + c). To find the product of two binomials, multiply the first term of each binomial, the outside terms, the inside terms, and the last terms. Then, add the products. (x + c)(x + c) = x2 + cx + cx + c2 = x2 + 2cx + c2.
  6. d.   To find the factors of a quadratic, begin by finding two numbers whose product is equal to the constant of the quadratic. Of those numbers, find the pair that adds to the coefficient of the x term of the quadratic. However, this quadratic has no x term; the sum of the products of the outside and inside terms of the factors is 0. –2 and 2 multiply to –4 and add to 0. Therefore, the factors of x2 – 4 are (x – 2) and (x + 2).
  7. e.   Begin with the innermost function: find f(3) by substituting 3 for x in the function f(x): f(3) = 2(3) + 1 = 6 + 1 = 7. Next, substitute the result of that function for x in g(x): g(7) = 7 – 2 = 5. Finally, substitute 5 for x in f(x): f(5) = 2(5) + 1 = 10 + 1 = 11.
  8. c.   Begin with the innermost function. You are given the value of f(x): f(x) = 6x + 4. Substitute this expression for x in the equation g(x): g(x) = x2 – 1; g(6x + 4) = (6x + 4)2 – 1 = 36x2 + 24x + 24x + 16 – 1 = 36x2 + 48x + 15. Therefore, g[f(x)] = 36x2 + 48x + 15.
  9. b.   The domain of a function is the set of all possible inputs to the function. All real numbers can be substituted for x in the function middle, excluding those that make the fraction undefined. Set the denominator equal to 0 to find the values of x that make the fraction undefined. These values are not in the domain of the function. x2 – 9 = 0, (x – 3)(x + 3) = 0, x – 3 = 0, x = 3; x + 3 = 0, x = –3. The domain of f(x) is all real numbers excluding 3 and –3.
  10. e.   The range of a function is the set of all possible outputs of the function. All real numbers can be substituted for x in the function f(x) = x2 – 4, so the domain of the function is all real numbers. Because the x term is squared, the smallest value that this term can equal is 0 (when x = 0). Therefore, the smallest value that f(x) can have is when x = 0. When x = 0, f(x) = 02 – 4 = –4. The range of f(x) is all real numbers greater than or equal to –4.
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