Please helppp!! Given the equation Square root of 2x plus 1 = 3, solve for x and identify if it is an extraneous solution. x = 4, solution is extraneous x = 4, solution is not extraneous x = 5, solution is extraneous x = 5, solution is not extraneous
\[\sqrt{2x}+1=3\]
\[\sqrt{2x+1}=3\] \[3^2=2x+1\] \[2x=8\] \[x=4\]
so its B because its not extraneous
but give medal to @Abmon98
I take no credit
@camerondoherty To be honest, I dont know what extraneous means.
extraneous means basically its FALSE and non extraneous means its TRUE with the solution i guess if you narrow it down
thats what my teacher told me and how i learned it
Thanks ! ♥
can any of you help me with one more question?
ask a new question and tag me in it like this @camerondoherty
Eloise started to solve a radical equation in this way: Square root of negative 2x plus 1 − 3 = x Square root of negative 2x plus 1 − 3 + 3 = x + 3 Square root of negative 2x plus 1 = x + 3 Square root of negative 2x plus 1 − 1 = x + 3 − 1 Square root of negative 2 x = x + 2 (Square root of negative 2 x)2 = (x − 4)2 −2x = x2 − 8x + 16 −2x + 2x = x2 + 8x + 16 + 2x 0 = x2 + 10x + 16 0 = (x + 2)(x + 8) x + 2 = 0 x + 8 = 0 x + 2 − 2 = 0 − 2 x + 8 − 8 = 0 − 8 x = −2 x = −8 Both solutions are extraneous because they don't satisfy the original equation. What error did Eloise make? She added 2x after squaring both sides. She subtracted 1 before squaring both sides. She factored x2 + 10x + 16 incorrectly. She did not check for extraneous solutions.
@camerondoherty idk how to take a picture of it so i had to put it here :/
can you link me the page ill take a puctiure
its on FLVS..
because i cant really understand that xD
just try it
lol okay let me see if i can hold on
@camerondoherty
oh gawsh idk sorry...
its okay thanks anyways !! :D
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