`sqrt(2x+6)+4 =x +3 `
What is the solution set of the equation above?

Options

A. {−1}
B. {5}
C. {−1, 5}
D. {0, −1, 5}

Answer

B

Explanation

Subtracting 4 from both sides of `sqrt(2x+6) + 4 = x+3` isolates the radical expression on the left side of the equation as follows: `sqrt(2X+6)` =x-1 . Squaring both sides of `sqrt(2X+6)`=x-1 yields `2x+6=x^2-2x+1` . This equation can be rewritten as a quadratic equation in standard form: `x^2 - 4x -5=0` . One way to solve this quadratic equation is to factor the expression `x^2 -4x -5` by identifying two numbers with a sum of –4 and a product of –5. These numbers are –5 and 1. So the quadratic equation can be factored as (x – 5)(x + 1) = 0. It follows that 5 and –1 are the solutions to the quadratic equation. However, the solutions must be verified by checking whether 5 and –1 satisfy the original equation, `sqrt(2X+6)+4 =x+3` .When X=-1 , the original equation gives , `sqrt(2-(-1)+6)+4 =(-1)+3` , or 6=2 , which is false. Therefore, –1 does not satisfy the original equation. When x = 5, the original equation gives `sqrt(2(5)+6)+4=5+3` , or 8=8 , which is true. Therefore, x = 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 `sqrt(2(0) + 6) + 4 = (0) + 3`, or `sqrt( 6) + 4 = 3`. This is not a true statement, so 0 is not a solution to the given equation.