what is the solution of $sqrt{1 - 3x}=x + 3$?\n$x=-8$ or $x = - 1$\n$x=-8$\n$x=-1$\nno solution

Explanation:

Step1: Square both sides

$(\sqrt{1 - 3x})^2=(x + 3)^2$ $1-3x=x^{2}+6x + 9$

Step2: Rearrange to quadratic form

$x^{2}+6x+3x+9 - 1=0$ $x^{2}+9x + 8=0$

Step3: Factor the quadratic

$(x + 8)(x+1)=0$

Step4: Solve for x

$x+8 = 0$ gives $x=-8$; $x + 1=0$ gives $x=-1$

Step5: Check for extraneous solutions

For $x=-8$, $\sqrt{1-3\times(-8)}=\sqrt{1 + 24}=5$, $-8 + 3=-5$, not a solution. For $x=-1$, $\sqrt{1-3\times(-1)}=\sqrt{4}=2$, $-1+3 = 2$, a solution.

Answer:

$x=-1$