JacksonJRB
  • JacksonJRB
Solve for x in terms of y. Please help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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JacksonJRB
  • JacksonJRB
\[y=\sqrt{\frac{ 4x-1 }{ x }}\]
JacksonJRB
  • JacksonJRB
Every new problem I look at seems to make less and less sense.
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle y^2=\frac{4x-1}{x} }\) \(\large\color{black}{ \displaystyle y^2=\frac{4x}{x}-\frac{1}{x} }\) \(\large\color{black}{ \displaystyle y^2=4-\frac{1}{x} }\) can you take it from there?

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SolomonZelman
  • SolomonZelman
all they are asking you to do, is to isolate the x (without performing incorrect operations).
JacksonJRB
  • JacksonJRB
\[x=\frac{ -1 }{ y^2-4 }\]
SolomonZelman
  • SolomonZelman
yes, and if you want you can simplify that just a bit, to get: \(\large\color{black}{ \displaystyle x=\frac{1}{4-y^2} }\)
JacksonJRB
  • JacksonJRB
Ah, thanks again
SolomonZelman
  • SolomonZelman
You are always welcome
JacksonJRB
  • JacksonJRB
There is another one that I will try on my own for now but I'll ask if I need help again
SolomonZelman
  • SolomonZelman
Ok:)
JacksonJRB
  • JacksonJRB
\[y=\frac{ 1 }{ x^2+2x }\] \[x^2+2x=\frac{ 1 }{ y }\]
SolomonZelman
  • SolomonZelman
the second equation is the next step that you took?
JacksonJRB
  • JacksonJRB
Yes
SolomonZelman
  • SolomonZelman
Ok, you can subtract -1/y from both sides, and do the quadratic formula. a=1 b=2 c=1/y
SolomonZelman
  • SolomonZelman
i mean c=-1/y
JacksonJRB
  • JacksonJRB
Ah, ok
JacksonJRB
  • JacksonJRB
ill try that
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x^2+2x\color{red}{-\frac{1}{y}}=\frac{1}{y}\color{red}{-\frac{1}{y}} }\) \(\large\color{black}{ \displaystyle x^2+2x-\frac{1}{y}=0 }\)
SolomonZelman
  • SolomonZelman
go on... :)
JacksonJRB
  • JacksonJRB
\[x=\frac{ -2+/-\sqrt{4+\frac{ 4 }{ y }} }{ }\]
JacksonJRB
  • JacksonJRB
over 2
SolomonZelman
  • SolomonZelman
you can do a \(\pm\) sign in latex by \pm (just a code advise) and yes, so far correct
JacksonJRB
  • JacksonJRB
\[x=-2\pm \sqrt{4y+4}\]
JacksonJRB
  • JacksonJRB
over 2 again
SolomonZelman
  • SolomonZelman
i don't think I followed you this time. How come your y came from denominator to numerator?
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x=\frac{-2\pm\sqrt{4-4(1)(-\frac{1}{y})}}{2} }\) \(\large\color{black}{ \displaystyle x=\frac{-2\pm\sqrt{4+4(\frac{1}{y})}}{2} }\) it should be like this
JacksonJRB
  • JacksonJRB
ah
SolomonZelman
  • SolomonZelman
it can be simplified ....
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle x=\frac{-2\pm\sqrt{4+4(\frac{1}{y})}}{2} }\) \(\large\color{black}{ \displaystyle x=\frac{-2\pm2\sqrt{1+(\frac{1}{y})}}{2} }\) \(\large\color{black}{ \displaystyle x=-1\pm\sqrt{1+\frac{1}{y}} }\)
SolomonZelman
  • SolomonZelman
this is it, technically, although, if you feel like or need to you can rationalize the denominator
JacksonJRB
  • JacksonJRB
Don't think do
JacksonJRB
  • JacksonJRB
Thank you so much
JacksonJRB
  • JacksonJRB
*think so
SolomonZelman
  • SolomonZelman
yw

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