We must solve the following equation for x:
[tex]x+3=\sqrt{3-x}[/tex]We can square both sides of the equation so we can get rid of the radical:
[tex]\begin{gathered} (x+3)^2=(\sqrt{3-x})^2 \\ (x+3)^2=3-x \end{gathered}[/tex]We expand the squared binomial on the left:
[tex]\begin{gathered} (x+3)^2=x^2+6x+9=3-x \\ x^2+6x+9=3-x \end{gathered}[/tex]Then we substract (3-x) from both sides:
[tex]\begin{gathered} x^2+6x+9-(3-x)=x-3-(3-x) \\ x^2+6x+9+x-3=0 \\ x^2+7x+6=0 \end{gathered}[/tex]Then we have to find the solutions to this last equation. Remember that the solutions to an equation of the form ax²+bx+c have the form:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]In our case a=1, b=7 and c=6 so we get:
[tex]\begin{gathered} x=\frac{-7\pm\sqrt{7^2-4\cdot1\cdot6}}{2\cdot1}=\frac{-7\pm\sqrt{49-24}}{2}=\frac{-7\pm\sqrt{25}}{2}=\frac{-7\pm5}{2} \\ x=\frac{-7+5}{2}=-1\text{ and }x=\frac{-7-5}{2}=-6 \end{gathered}[/tex]So we have two potential solutions x=-1 and x=-6. However we should note something important, in the original equation we have the term:
[tex]\sqrt{3-x}[/tex]Remember that the result of the square root is always positive. Then the term in the left of the expression has to be positive or 0. Then we impose a restriction in the value of x:
[tex]x+3\ge0\rightarrow x\ge-3[/tex]From the two possible solutions only x=-1 is greater than or equal to -3 so this is the correct one.
AnswerThen the answer is option A.
