Respuesta :
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]
Answer:
The actual answer is C. x = -1
Step-by-step explanation:
Remove the radical by raising each side to the index of the radical. x = − 1
<3 thank you expert verified for trying!