Destinyyyy
  • Destinyyyy
Solve the equation by making an appropriate substitution.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Destinyyyy
  • Destinyyyy
x^-2 +3x^-1 -10=0
freckles
  • freckles
let u=x^(-1) then u^2=x^(-2)
Destinyyyy
  • Destinyyyy
okay

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Destinyyyy
  • Destinyyyy
Do i factor now?
freckles
  • freckles
yep yep
freckles
  • freckles
and you should be factoring u^2+3u-10 right now
Destinyyyy
  • Destinyyyy
(u-2)(u+5)
freckles
  • freckles
neat so you can solve the equation (u-2)(u+5)=0 for u
Destinyyyy
  • Destinyyyy
u=2, -5
freckles
  • freckles
now remember we replace x^(-1) with u so we have to undo that replacement by replacing u with x^(-1)
freckles
  • freckles
\[x^{-1}=2 \text{ or } -5 \\ x^{-1}=2 \text{ or } x^{-1}=-5 \]
freckles
  • freckles
recall x^(-1) is the same thing as saying 1/x
freckles
  • freckles
\[\frac{1}{x}=2 \text{ or } \frac{1}{x}=-5 \]
freckles
  • freckles
can you solve both of those equations for x
Destinyyyy
  • Destinyyyy
?? x^-2=2 you have two -1
freckles
  • freckles
\[\frac{1}{x}=a \\ \text{ multiply both sides by } x \\ 1=ax \\ \text{ divide both sides by } a \\ \frac{1}{a}=x \\ \text{ so assuming } x \neq 0 \text{ and } a \neq 0 \\ \text{ then } \frac{1}{x}=a \implies x=\frac{1}{a}\]
Destinyyyy
  • Destinyyyy
Um okay I think I understand that but dont see how that solves this.. :/
Destinyyyy
  • Destinyyyy
Dont I just square root both sides?
freckles
  • freckles
you have 1/x=2 so x=1/2
freckles
  • freckles
you have 1/x=-5 so x=?
Destinyyyy
  • Destinyyyy
-1/5
freckles
  • freckles
right
Destinyyyy
  • Destinyyyy
So thats the answer?
freckles
  • freckles
x=1/2 or x=-1/5 yes
Destinyyyy
  • Destinyyyy
Do I check it? Or no?
freckles
  • freckles
another way to solve this is multiply both sides of your equation by x^2 \[x^{-2}+3x^{-1}-10=0 \\ x^2(x^{-2}+3x^{-1}-10)=x^{2}(0) \\ x^{2+(-2)}+3x^{2+(-1)}-10x^{2}=0 \\ x^{0}+3x^{1}-10x^2=0 \\ -10x^2+3x+1=0 \text{ since } x^{0}=1 \text{ and I just reorder terms } \\ -10x^2+5x-2x+1=0 \\ -5x(2x-1)-1(2x-1)=0 \\ (2x-1)(-5x-1)=0 \\ 2x-1=0 \text{ when } x=\frac{1}{2} \\ -5x-1=0 \text{ when } x=\frac{-1}{5}\] there is no harm in checking but the answers are fine we didn't raise both sides to an even power and are answers fit in with domain of the original equation (for this question we definitely did not want x=0 as a solution which we didn't have to worry about here)
freckles
  • freckles
oh I see what you did about you replace u with x^(-2) even though u equals x^(-1)
Destinyyyy
  • Destinyyyy
Um okay
freckles
  • freckles
above not about :p
freckles
  • freckles
anyways if you have more questions post a new question I'm gonna take a short break
freckles
  • freckles
unless you have questions on this one then I can stay and answer them real quick
freckles
  • freckles
ok I will be back later! peace for now! :)
Destinyyyy
  • Destinyyyy
Sorry didnt see you send back.

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