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Yields 2 x plus 6 equals x squared minus 2 x plus 1

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yields 2 x plus 6, equals x squared, minus 2 x , plus 1 . This equation can be rewritten as a quadratic equation in standard form: x squared, minus 4 x , minus 5, equals 0 . One way to solve this quadratic equation is to factor the expression x squared, minus 4 x , minus 5 by identifying two numbers with a sum of negative 4 and a product of negative 5 . These numbers are negative 5 and 1. So the quadratic equation can be factored as open parenthesis, x minus 5, close parenthesis, times, open parenthesis, x plus 1, close parenthesis, equals 0 . It follows that 5 and negative 1 are the solutions to the quadratic equation. However, the solutions must be verified by checking whether 5 and negative 1 satisfy the original equation, the square root of 2 x plus 6, end root, plus 4, equals, x plus 3 . When x equals negative 1 , the original equation gives the square root of 2 times negative 1, plus 6, end root, plus 4, equals, negative 1 plus 3 , or 6 equals 2 , which is false. Therefore, negative 1 does not satisfy the original equation. When x equals 5 , the original equation gives the square root of 2 times 5, plus 6, end root, plus 4, equals, 5 plus 3 , or 8 equals 8 , which is true. Therefore, x equals 5 is the only solution to the original equation, and so the solution set is 5 . Incorrect answer Page 6
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Choices A, C, and D are incorrect because each of these sets contains at least one value that results in a false statement when substituted into the given equation. For instance, in choice D, when 0 is substituted for x into the given equation, the result is the square root of 2 times 0, plus 6, end root, plus 4, equals, 0 plus 3 , or the square root of 6, end root, plus 4, equals 3 . This is not a true statement, so 0 is not a solution to the given equation. Page 7
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Explanation for question 8. Correct answer Choice D is correct. Since x cubed, minus 9 x , equals, x times, open parenthesis, x plus 3, close parenthesis, times, open parenthesis, x minus 3, close parenthesis and x squared, minus 2 x , minus 3, equals, open parenthesis, x plus 1, close parenthesis, times, open parenthesis, x minus 3, close parenthesis , the fraction f of x , over g of x can be written as the fraction with numerator x times, open parenthesis, x plus 3, close parenthesis, times, open parenthesis, x minus 3, close parenthesis, and denominator, open parenthesis, x plus 1, close parenthesis, times, open parenthesis, x minus 3, close parenthesis, end fraction . It is given that x is greater than 3 , so the common factor x minus 3 is not equal to 0. Therefore, the fraction can be further simplified to the fraction with numerator x times, open parenthesis, x plus 3, close parenthesis, and denominator x plus 1, end fraction . Incorrect answer Page 8
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Choice A is incorrect. The expression the fraction 1 over, x plus 1, end fraction is not equivalent to the fraction f of x , over g of x because at x equals 0 , the fraction 1 over, x plus 1, end fraction as a value of 1 and the fraction f of x , over g of x has a value of 0. Choice B is incorrect and results from omitting the factor x in the factorization of f of x . Choice C is
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