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Mathematics 15 Online
OpenStudy (destinyyyy):

Solve the equation by making an appropriate substitution.

OpenStudy (destinyyyy):

x^-2 +3x^-1 -10=0

OpenStudy (freckles):

let u=x^(-1) then u^2=x^(-2)

OpenStudy (destinyyyy):

okay

OpenStudy (destinyyyy):

Do i factor now?

OpenStudy (freckles):

yep yep

OpenStudy (freckles):

and you should be factoring u^2+3u-10 right now

OpenStudy (destinyyyy):

(u-2)(u+5)

OpenStudy (freckles):

neat so you can solve the equation (u-2)(u+5)=0 for u

OpenStudy (destinyyyy):

u=2, -5

OpenStudy (freckles):

now remember we replace x^(-1) with u so we have to undo that replacement by replacing u with x^(-1)

OpenStudy (freckles):

\[x^{-1}=2 \text{ or } -5 \\ x^{-1}=2 \text{ or } x^{-1}=-5 \]

OpenStudy (freckles):

recall x^(-1) is the same thing as saying 1/x

OpenStudy (freckles):

\[\frac{1}{x}=2 \text{ or } \frac{1}{x}=-5 \]

OpenStudy (freckles):

can you solve both of those equations for x

OpenStudy (destinyyyy):

?? x^-2=2 you have two -1

OpenStudy (freckles):

\[\frac{1}{x}=a \\ \text{ multiply both sides by } x \\ 1=ax \\ \text{ divide both sides by } a \\ \frac{1}{a}=x \\ \text{ so assuming } x \neq 0 \text{ and } a \neq 0 \\ \text{ then } \frac{1}{x}=a \implies x=\frac{1}{a}\]

OpenStudy (destinyyyy):

Um okay I think I understand that but dont see how that solves this.. :/

OpenStudy (destinyyyy):

Dont I just square root both sides?

OpenStudy (freckles):

you have 1/x=2 so x=1/2

OpenStudy (freckles):

you have 1/x=-5 so x=?

OpenStudy (destinyyyy):

-1/5

OpenStudy (freckles):

right

OpenStudy (destinyyyy):

So thats the answer?

OpenStudy (freckles):

x=1/2 or x=-1/5 yes

OpenStudy (destinyyyy):

Do I check it? Or no?

OpenStudy (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)

OpenStudy (freckles):

oh I see what you did about you replace u with x^(-2) even though u equals x^(-1)

OpenStudy (destinyyyy):

Um okay

OpenStudy (freckles):

above not about :p

OpenStudy (freckles):

anyways if you have more questions post a new question I'm gonna take a short break

OpenStudy (freckles):

unless you have questions on this one then I can stay and answer them real quick

OpenStudy (freckles):

ok I will be back later! peace for now! :)

OpenStudy (destinyyyy):

Sorry didnt see you send back.

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