SAT Question 66bce0c1: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice B is correct. 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 $\{5\}$.