Answer:
x = -1
Step-by-step explanation:
[tex]\sqrt{1-3x}=x+3\\\\1-3x=x^2+6x+9 \qquad\text{square both sides}\\\\x^2+9x+8=0 \qquad\text{put in standard form}\\\\(x+8)(x+1)=0 \qquad\text{factor}\\\\\left \{ {{x=-8} \atop {x=-1}} \right. \qquad\text{values that make the factors zero}[/tex]
Only the solution x = -1 will work for this equation. The other solution is extraneous.
