What is the solution of StartRoot 1 minus 3 x EndRoot = x + 3 ?x = –8 or x = –1x = –8x = –1no solution

Answer:[tex]x=-8[/tex] or [tex]x=-1[/tex]Step-by-step explanation:We have the following equation:[tex]\sqrt{1-3x}=x+3[/tex]The first step is to eliminate the square root, it moves to the right side as a power of two:[tex]1-3x=(x+3)^2[/tex]Now, we develop the right side as a squared binomial:[tex]1 -3x=x^2+6x+9[/tex]we put everything together on one side of the equation:[tex]x^2+6x+9-1+3x=0[/tex]We join like terms:[tex]x^2+9x+8=0[/tex]and we factor the previous equation. For this we will look for two numbers such that:when multiplying they give us 8, and when they are added they give us 9. those numbers are 8 and 1, since 8*1=8 and 8+1=9.So the factorization will be as follows:[tex](x+8)(x+1)=0[/tex]and the above can have two results, that the first parenthesis is zero, or that the second parenthesis is zero:[tex](x+8)=0\\x=-8[/tex]or [tex](x+1)=0\\x=-1[/tex]The answer fot the equation is [tex]x=-8[/tex] or [tex]x=-1[/tex]