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Advanced Math / Nonlinear equations in one variable and systems of equations in two variables Difficulty: Hard

$The square root of 2 x plus 6, end root, plus 4 equals, x plus 3$

What is the solution set of the equation above?

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Explanation

Choice B is correct. Subtracting 4 from both sides of $the square root of 2 x plus 6, end root, plus 4, equals, x plus 3$ isolates the radical expression on the left side of the equation as follows: $the square root of 2 x plus 6, end root, equals, x minus 1$ . Squaring both sides of $the square root of 2 x plus 6, end root, equals, x minus 1$ 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 $5$ .

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 $the square root of 2 times 0, plus 6, plus 4, end root, equals, 0 plus 3$ , 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.