Solve the equation by making an appropriate substitution.
x^-2 +3x^-1 -10=0
let u=x^(-1) then u^2=x^(-2)
okay
Do i factor now?
yep yep
and you should be factoring u^2+3u-10 right now
(u-2)(u+5)
neat so you can solve the equation (u-2)(u+5)=0 for u
u=2, -5
now remember we replace x^(-1) with u so we have to undo that replacement by replacing u with x^(-1)
\[x^{-1}=2 \text{ or } -5 \\ x^{-1}=2 \text{ or } x^{-1}=-5 \]
recall x^(-1) is the same thing as saying 1/x
\[\frac{1}{x}=2 \text{ or } \frac{1}{x}=-5 \]
can you solve both of those equations for x
?? x^-2=2 you have two -1
\[\frac{1}{x}=a \\ \text{ multiply both sides by } x \\ 1=ax \\ \text{ divide both sides by } a \\ \frac{1}{a}=x \\ \text{ so assuming } x \neq 0 \text{ and } a \neq 0 \\ \text{ then } \frac{1}{x}=a \implies x=\frac{1}{a}\]
Um okay I think I understand that but dont see how that solves this.. :/
Dont I just square root both sides?
you have 1/x=2 so x=1/2
you have 1/x=-5 so x=?
-1/5
right
So thats the answer?
x=1/2 or x=-1/5 yes
Do I check it? Or no?
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)
oh I see what you did about you replace u with x^(-2) even though u equals x^(-1)
Um okay
above not about :p
anyways if you have more questions post a new question I'm gonna take a short break
unless you have questions on this one then I can stay and answer them real quick
ok I will be back later! peace for now! :)
Sorry didnt see you send back.
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