Question

What is the solution of StartRoot 1 minus 3 x EndRoot = x + 3 ?

Answer 1

First square both sides

1-3x=(x+3)^2 use foil to unfactor it
1-3x=x^2+6x+9 subtract 1 from both sides
-3x=x^2+6x+8 add 3x to both sides
0=x^2+9x+8

Now that we got it equal to zero you have to factor the new trinomial

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

Now since both are being multiplied by each other, if you get 1 of them to equal zero then both equal zero since 0 times anything equals 0.

So your solutions are x=-8 and x=-1

Brainliest my answer if it helps you out?

Answer 2

Answer:

$x=-8$ or $x=-1$

Step-by-step explanation:

We have the following equation:

$\sqrt{1-3x}=x+3$

The first step is to eliminate the square root, it moves to the right side as a power of two:

$1-3x=(x+3)^2$

Now, we develop the right side as a squared binomial:

$1 -3x=x^2+6x+9$

we put everything together on one side of the equation:

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

We join like terms:

$x^2+9x+8=0$

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:

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

and the above can have two results, that the first parenthesis is zero, or that the second parenthesis is zero:

$(x+8)=0\\x=-8$

or

$(x+1)=0\\x=-1$

The answer fot the equation is $x=-8$ or $x=-1$