Question

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

Answer 1

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 soluti

Answer 2

@jdoe0001

Answer 3

you don't happen to have a screenshot of it?

Answer 4

hey you go

Answer 5

\(\bf \sqrt{-2x+1}-3=x\\
\sqrt{-2x+1}-3+3=x+3\\
\sqrt{-2x+1}=x+3\\
\color{red}{\sqrt{-2x+1}-1=x+3-1}\\
\sqrt{-2x}=x+2\\
(\sqrt{-2x})^2=(x+2)^2\\
\color{red}{-2x = x^2-8x+16\\
-2x+2x=x^2+8x+16+2x}\\
0=x^2+10x+16\\
0=(x+2)(x+8)\)

do you notice anything in those lines?

Answer 6

they add wrong?

Answer 7

hmm actually even more

\(\bf \bf \sqrt{-2x+1}-3=x\\
\sqrt{-2x+1}-3+3=x+3\\
\sqrt{-2x+1}=x+3\\
\color{red}{\sqrt{-2x+1}-1=x+3-1}\\
\color{blue}{\sqrt{-2x}=x+2}\\
\color{red}{(\sqrt{-2x})^2=(x-4)^2}\\
\color{blue}{-2x = x^2-8x+16}\\
\color{red}{-2x+2x=x^2+8x+16+2x}\\
0=x^2+10x+16\\
0=(x+2)(x+8)\)

Answer 8

\(\bf \sqrt{-2x+1}-1=x+3-1\qquad \textit{can't substract from a radical}\\
\textit{unless the number is inside the radical already}\)

Answer 9

now as far as the others, I gather are typos, notice in the 1st blue line to the red line

it goes from x+2 to x-4 for whatever reason

notice the 2nd blue to the red line

it goes from \(\bf x^2-8x+16 \ \ to \ \ x^2+8x+16\) changing sign on the 8x

Answer 10

so would it be he factored x2 + 10x + 16 incorrectly.

Answer 11

\(\bf \large {\sqrt{-2x+1}-1\implies \sqrt{-2x+1}-1\\ \quad \\
\sqrt{-2x+1-1}\implies \sqrt{-2x}\\ \quad \\
\sqrt{-2x+1}-1\ne \sqrt{-2x}}\)

Answer 12

hmm if you do FOIL on (x+2)(x+8) it will give you \(\bf x^2+10x+16\)

Answer 13

so that's factored ok

Answer 14

im so confused would it be She did not check for extraneous solution?

Answer 15

she wanted to isolate the radical and -2x in it, and she wanted to rid of the +1
for that you'd have to square both sides, so you can take out the amount in the radical AND THEN you remove the +1

she used -1 to both sides prematurely, and incorrectly

Answer 16

o i kind of of get it

Answer 17

\(\large \begin{array}{llll}
\textit{what she did}&\qquad &\textit{what she should have done}\\
\hline\\
\sqrt{-2x+1}-1=x+3-1&& (\sqrt{-2x+1})^2=(x+3)^2\\
\sqrt{-2x}=x+2&& -2x+1=(x+3)^2\\
&&-2x+1-1=(x+3)^2-1\\
&&-2x = (x+3)^2-1
\end{array}\)