Question

3x^(-2)+2x^(-1)-8=0

Answer 1

@John_ES

Answer 2

put \(u=x^{-1}\) and solve \[3u^2+2u-8=0\]

Answer 3

once you have \(u\) replace it by \(x^{-1}\) which means \(\frac{1}{x}\) and solve

Answer 4

this one is very much like the last one you had
sometimes they come under the heading of "equations of quadratic type" because you can make them look like quadratic equations
if you want me to check your answer for \[3u^2+2u-8=0\] i can

Answer 5

@satellite73 i currently have -2+-10/6

somehow the answer is -1/2 and 3/4.. Im confused.

Answer 6

ok let me check

Answer 7

the answer is \[u=-2,u=\frac{5}{3}\] right? i mean after you reduce

Answer 8

dang typo, i meant \[y=-2,y=\frac{4}{3}\] or did you use the quadratic formula?

Answer 9

my answr key says -1/2 and 3/4... is it wrong?

Answer 10

no it isright, we are not done yet

Answer 11

You need to return to your original variable x, as in the problem we did before.

Answer 12

\[u=-2\] but \(u=\frac{1}{x}\) so if \[\frac{1}{x}=-2\] then \[x=-\frac{1}{2}\] i.e the reciprocal

Answer 13

so when i have -2+-10/6, how do i get to -2 and 4/3??

Answer 14

oho i see you used the quadratic formula, like i though

Answer 15

you actually have two numbers there \[\frac{-2+10}{6}\] and \[\frac{-2+10}{6}\]

Answer 16

oops one with a minus sign \[\frac{-2-10}{6}\]

Answer 17

now do you see the \(-2\)?

Answer 18

\[\frac{-2-10}{6}=\frac{-12}{6}=-2\]

Answer 19

yes, okay

Answer 20

i also see the 4/3

Answer 21

right good

Answer 22

so know we have \[u=-2\\
u=\frac{4}{3}\]

Answer 23

therefore \[x^{-1}=-2\\
x^{-1}=\frac{4}{3}\]

Answer 24

but \(x^{-1}=\frac{1}{x}\) so we really have \[\frac{1}{x}=-2\\
\frac{1}{x}=\frac{4}{3}\]

Answer 25

awesome! that makes sense. thank you :))))

Answer 26

to solve for \(x\) flip both sides
you good now?

Answer 27

yes