the square root of 2x +3 =x would the answer be 3 only or -1 and 3?

Answer 1

I assume you mean sq rt (2x + 3) =x
{sqrt(2x+3)}^2 =x^2
2x+3=x^2
X^2 -2x -3 = 0
(x -3)(x+1)=0

Now, for (x-3)(x+1) to multiply out to zero, one or both of those factors have to be zero. We don't know which, so we say
x-3=0, OR x+1=0
If it's x-3=0, x=3. If it's x+1=0, x=-1

YOU HAVE TO SUBSTITUTE BOTH ANSWERS
BACK INTO THE ORIGINAL EQUATION

Sorry, didn't mean to shout, but this step is the key to your answer.
If x=3, sqrt(2x+3) =3 becomes sqrt9 =3, 3=3
If x=-1,sqrt(2x+3)=3 becomes sqrt(-2+3)=3, 1=3.
Doesn't work.
Your only acceptable answer is x=3.

This happens because when you square both sides, any negative terms become positive, and therefore can mislead you. Whenever squaring a sqrt term is involved you always have to check your answers by
substituting back into the ORIGINAL equation.

Hope this helps. Good luck!

Answer 2

Hello

Square both sides and we have 2x+3 = x^2 now put everything on the right side giving us 0 = x^2 -2x -3 factoring gives 0 = (x - 3) (x +1) so we have x = 3 and x = -1. Now put these into the original to see if they check.
sqrt(2*3 + 3) = sqrt 9 = 3 this one checks
sqrt(2*(-1) + 3) = 1 not -1This does not check.

Hope This Helps!!

Answer 3

x = -3

Answer 4

just 3

Answer 5

Try plugging the -1 into your original.
You would get sq rt of (2*(-1) + 3) which is 1.
On the right you would have -1.
1 = -1 is not a true statement therefore -1 is not a solution.

Answer 6

2x +3 = x ---> x = -3

sqrt(-3) = i * sqrt(3) where i is the imaginary number = sqrt(-1)