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!

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