x3_drummerchick
  • x3_drummerchick
can someone explain to me how my teacher got this answer? its using substitution. Will give medals! give me one sec to post the problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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x3_drummerchick
  • x3_drummerchick
\[x^{1/2}+3x^{-1/2}-54x^{-3/2}=0\]
x3_drummerchick
  • x3_drummerchick
solve for x
anonymous
  • anonymous
\[\sqrt{x}+\frac{ 3 }{ \sqrt{x} }-\frac{ 54 }{ \left( \sqrt{x} \right)^3 }=0\] \[put~\sqrt{x}=t\] \[t+\frac{ 3 }{ t }-\frac{ 54 }{ t^3 }=0\] \[multiply~ by ~t^3\] \[t^4+3t^2-54=0\] put t^2=y \[y^2+3y-54=0\] \[y^2+9y-6y-54=0\] y=? then t=? x=?

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x3_drummerchick
  • x3_drummerchick
my teacher told us to try it like this: @surjithayer
x3_drummerchick
  • x3_drummerchick
i dont understand the way he showed us in the above file
Nnesha
  • Nnesha
yea i was abt to solve like ^^that (Factor )
x3_drummerchick
  • x3_drummerchick
where did the y come from?
x3_drummerchick
  • x3_drummerchick
our teacher wants us to factor it
x3_drummerchick
  • x3_drummerchick
i just dont get how he pulled out that exponential fraction
Nnesha
  • Nnesha
but i also like the way surjithayer's solved
x3_drummerchick
  • x3_drummerchick
i need to learn how to factor out the negative exponent :(
Nnesha
  • Nnesha
he supposed that t^2 = y
Nnesha
  • Nnesha
so t^4 is same as t^2 times t^2 replace t^2 with y y times y = y^2 :=)
x3_drummerchick
  • x3_drummerchick
i know how to use the substitution method, im just a little confused with how x^-1/2 was fatored out.
Nnesha
  • Nnesha
x^{-1/2) or x^{-3/2} ?
x3_drummerchick
  • x3_drummerchick
x^-3/2, sorry
Nnesha
  • Nnesha
what's the lowest exponent of x ? that would be your common factor
Nnesha
  • Nnesha
wait i'll give you an example 2x^2+3x <--x^1 degree would be common factor
x3_drummerchick
  • x3_drummerchick
okay gotcha
Nnesha
  • Nnesha
so what is common factor in ur equation ?
x3_drummerchick
  • x3_drummerchick
x^-3/2, ause its the smallest
Nnesha
  • Nnesha
yes right!\[\huge\rm x^\frac{ 1 }{ 2 }+ 3x^\frac{ -1 }{ 2}+54x^\frac{ -3 }{ 2 }\] take out the common factor when you take out x^{-3/2} from x^{1/2} what will ou have left ?\[x^\frac{ -3 }{ 2}(??????????????)\]
x3_drummerchick
  • x3_drummerchick
i dont know
Nnesha
  • Nnesha
in other words divide all 3 terms by common factor \[\huge\rm \frac{ x^\frac{ 1 }{ 2 } }{ x^\frac{ -3 }{ 2 } }+\frac{ 3x^\frac{ -1 }{ 2 } }{ x^\frac{ -3 }{ 2 } }+\frac{ 54x^\frac{ -3 }{ 2 } }{ x^\frac{ -3 }{ 2 }}\]
Nnesha
  • Nnesha
remember when we divide same bases we should `subtract `their exponents (exponent rule) \[ \huge\rm \frac{ x^m }{ x^n }=x^{m-n}\]
x3_drummerchick
  • x3_drummerchick
^ thats ringing a bell
Nnesha
  • Nnesha
:P
Nnesha
  • Nnesha
example \[\huge\rm \frac{ x^2 }{ x^1}=x^{2-1}\]
x3_drummerchick
  • x3_drummerchick
thats making such more sense
Nnesha
  • Nnesha
\[\huge\rm \color{ReD}{ \frac{ x^\frac{ 1 }{ 2 } }{ x^\frac{ -3 }{ 2 } }}+\frac{ 3x^\frac{ -1 }{ 2 } }{ x^\frac{ -3 }{ 2 } }+\frac{ 54x^\frac{ -3 }{ 2 } }{ x^\frac{ -3 }{ 2 }}\] look at first term bases are the same so subtract their exponents \[\huge\rm \color{Red}{x^{\frac{ 1 }{ 2 }-(\frac{ -3 }{ 2 })}}\]
x3_drummerchick
  • x3_drummerchick
i think i get it now, ou have to diivide all by it, that reduces it to the ax^2+bx+c
Nnesha
  • Nnesha
ye!!!
x3_drummerchick
  • x3_drummerchick
wow youre good, thanks!!!!
Nnesha
  • Nnesha
\[\huge\rm x^\frac{ -3 }{ 2 }(????+???+??)\] write ur answer in the parentheses
x3_drummerchick
  • x3_drummerchick
(x^2+3x-54)
Nnesha
  • Nnesha
yep right!
x3_drummerchick
  • x3_drummerchick
ive got it from this point out. shew you have no idea how long ive been trying to decifer this shenninagans!
Nnesha
  • Nnesha
aww se now you got it! great job!
x3_drummerchick
  • x3_drummerchick
thank you so much

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