Question

SOLVE
(2x-5/(x-2)^2=3/(x+2)

Answer 1

\[(2x-5/(x-2)^2=3/(x+2)\]

Answer 2

ok well i would distribute the exponent first

Answer 3

Ok

Answer 4

that gives you this correct? (2x−5/(x−2)(2x−5/(x−2)=3/(x+2)

Answer 5

yes

Answer 6

now get rid of the denominator

Answer 7

what do you get?

Answer 8

Idk? That's why i'm asking for help.

Answer 9

first off where does your missing parenthesis go?

Answer 10

What?

Answer 11

No. Oh so the 2x-5 is over x-2 and all of that has an exponent of two. And then that is equal to 3 over x+2

Answer 12

so (2x−5/(x−2))^2=3/(x+2)

Answer 13

icey, the way you wrote down the problem is confusing :) it's missing something or something is not there :)

Answer 14

sorry ((2x−5)/(x−2))^2=3/(x+2)

Answer 15

Nick, that's wha tI wrote down. and i don't think anything is missing.

Answer 16

that minus a couple parenthesis which makes a difference

Answer 17

is it [2x-5/(x-2)]^2=3/(x+2)

or 2x-5/(x-2)^2=3/(x+2)

Answer 18

unless your in a really advanced math class i think you messed something up

Answer 19

No Nick you had it correct.;

Answer 20

x~~-1.42875-1.11022x10^-16 i
x~~2.35334+2.22045x10^-16 i
x~~2.82541-2.22045x10^-16 i
these are your answers so i think you messed something up

Answer 21

2x-5/(x-2)^2=3/(x+2)
which is the correct equation ??? :
\(\large \frac{2x-5}{(x-2)^2}=\frac{3}{x+2} \)
or is it:
\(\large (\frac{2x-5}{x-2})^2=\frac{3}{x+2} \)

Answer 22

second one

Answer 23

yeah, thank you byteme :) i wish i know how to write like you do.

Answer 24

since it's the second one then:
\(\large (\frac{2x-5}{x-2})^2=\frac{3}{x+2} \)
\(\large \frac{(2x-5)^2}{(x-2)^2}=\frac{3}{x+2} \)
\(\large (2x-5)^2(x+2)^2=3(x-2)^2 \) just "cross multiply"

yuck... that left side is a 4th degree polynomial....

Answer 25

my bad... actually a 3rd degree polynomial. it should be:
\(\large (2x-5)^2(x+2)=3(x-2)^2 \)

Answer 26

2x^2+4x-5x-10=3x^2+12+6x =>
x^2+7x+2 => X1=(-7V41)/2 or X2=(-7V42)/2