Question what-i

2.2 Nonlinear Equations in One Variable - Quadratic solving (factoring, quadratic formula, completing square)
0:00

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?

A.

set consisting of negative 1

B.

set consisting of 5

C.

set consisting of negative 1 and 5

D.

set consisting of 0, negative 1, and 5

What is the solution set of the equation above?

Hard-difficulty · SAT Math · Nonlinear Equations in One Variable — Quadratic solving (factoring, quadratic formula, completing square). Read the question above, select your answer, and check the full explanation below to understand exactly why the correct choice works.

Answer 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 andnegative 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.