Question

???? 3x^-2+2x^-1-8=0
How do i find all solutions to this equation?

Answer 1

i've never worked with negative exponents in an equation like this and dont know how to begin working with it.

Answer 2

There are two ways to solve this question:
first) move the constant term to the right side and then take out the common factor
2nd) change negative exponent to positive exponents and then find the common denominator

Answer 3

Isn't the constant term and common factor the same thing in this scenario?

Answer 4

Multiply both sides by x^2, that'll positivize all the exponents.

Answer 5

Hmm i don't think so.
the constant term isn't the common factor.

Answer 6

*multiply by x^2* that is a good approach suggested by agent
one step and then you will get the quadratic equation.
go with that.

Answer 7

put \[x ^{-1}=y,squaring ~,\left( x ^{-1} \right)^2=y^2~or~x ^{-2}=y^2\]
\[3y^2+2y-8=0\]
\[3y^2+6y-4y-8=0\]
make factors and find y
then
\[y=x ^{-1}=\frac{ 1 }{ x },~or~x=\frac{ 1 }{ y }\]

Answer 8

\[\frac{3}{x^2}+\frac{2}{x}-8=0 \]\[\left\{x=-\frac{1}{2},x=\frac{3}{4}\right\} \]

Answer 9

@robtobey so initially dividing the terms by their power and then solving for x works? all the same?

Answer 10

He didn't divide the terms by their power... \[\Large x^{-a} = \frac{ 1 }{ x^a }\]