Question

solve each equation. Check your answers and identify any extraneous roots.

x - 2/x^2 - x - 6 = 1/x^2-4 + 3/2x + 4=

Answer 1

@ganeshie8 think u can help?

Answer 2

is the question pasted correctly

Answer 3

= in the end is throwing me off

Answer 4

ignore that equal sign.

Answer 5

i had to write it out cause its on paper. -.-

Answer 6

\(\large \frac{ x - 2}{x^2 - x - 6} = \frac{1}{x^2-4} + \frac{3}{2x + 4}\)

Answer 7

in your paper, is the question looking as above ?

Answer 8

yess

Answer 9

this requires lot of paper !

Answer 10

ya it looks like it..lol

Answer 11

ok lets do it

Answer 12

do you knw factoring ?

Answer 13

not really i never took algebra 2 in a classroom so i dont know anything :/

Answer 14

its okay factoring is easy

Answer 15

lets start wid factoring bottom :- x^2-x-6

Answer 16

we need to write that as product of two things. thats called factoring

Answer 17

oh ok

Answer 18

this is how we going to factor it :

x^2 - x - 6

x^2 -3x + 2x -6

x(x-3) + 2(x-3)

(x-3)(x+2)

Answer 19

make sure you understand how we factored, we going to do the same for many future problems. so you need to understand them if u want to feel comfortable when solving other problems later. ok. so just make sure u understand how to factor. if u dont, just ask ok

Answer 20

take ur time

Answer 21

ok how did you get 3x & the extra 2x?

Answer 22

good q, didnt u notice x is missing

Answer 23

x^2 - x - 6

x^2 -3x + 2x -6

Answer 24

yes & also the addition symbol

Answer 25

we need to rewrite the middle "-x"

Answer 26

i wrote, "-3x + 2x" in place of "-x"

Answer 27

il tell why we picked those soon, but do u agree that there is nothing wrong in what we did ?

Answer 28

cuz, -3x + 2x = -x

Answer 29

yes i seee

Answer 30

ok, lets move on. (il tell you how to pick the numbers later)

Answer 31

x^2 - x - 6

x^2 -3x + 2x -6

Answer 32

ok sounds good!

Answer 33

from the first two terms, take x common

Answer 34

x^2 - x - 6

x^2 -3x + 2x -6

x(x-3) + 2x - 6

Answer 35

still wid me ?

Answer 36

from the last two terms take 2 common

Answer 37

x^2 - x - 6

x^2 -3x + 2x -6

x(x-3) + 2x - 6

x(x-3) + 2(x-3)

Answer 38

now we see (x-3) is common in both terms, so take it common !

Answer 39

x^2 - x - 6

x^2 -3x + 2x -6

x(x-3) + 2x - 6

x(x-3) + 2(x-3)

(x-3)(x+2)

Answer 40

ok im kinda with you :p

Answer 41

we are done factoring ! just go thru and make yourself convinced u understand

Answer 42

so bottom x^2-x-6 can be written in factored form as (x-3)(x+2).

lets go to our main equation

Answer 43

okaay

Answer 44

\(\large \frac{ x - 2}{x^2 - x - 6} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

Answer 45

\(\large \frac{ x - 2}{x^2 - x - 6} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

Answer 46

next look at right side of equation. lets keep focusing on bottom

Answer 47

we have x^2-4

Answer 48

i want to factor it.

Answer 49

can u help me if u have any ideas how to factor it ?

Answer 50

lemme try

Answer 51

heres the clue :- write 4 as 2^2

and use the formula, a^2-b^2 = (a+b)(a-b)

Answer 52

ok il help you wid this

Answer 53

x^2 - 4

x^2 - 2^2

(x+2)(x-2)

Answer 54

we used a^2-b^2 formula, when moving from 2nd step to 3rd step

Answer 55

we have our main working as :-

\(\large \frac{ x - 2}{x^2 - x - 6} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2x + 4} \)

Answer 56

look at the bottom of last term.

2x+4

2(x+2)

Answer 57

\(\large \frac{ x - 2}{x^2 - x - 6} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2(x + 2)} \)

Answer 58

if you see (x+2) is there in the bottom for each term. so it can be cancelled

Answer 59

\(\large \frac{ x - 2}{x^2 - x - 6} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2(x + 2)} \)

\(\large \frac{ x - 2}{(x-3)\cancel{(x+2)}} = \frac{1}{\cancel{(x+2)}(x-2)} + \frac{3}{2\cancel{(x + 2)}} \)

\(\large \frac{ x - 2}{(x-3)} = \frac{1}{(x-2)} + \frac{3}{2} \)

Answer 60

you still wid me or i lost u lol, this is a very lengthy prob. but simple one.

Answer 61

ya im with you im just trying to write it out too to understand it >.<

Answer 62

good :) thats the only simplification possible by factoring. now we need to do the donkey work. get ready

Answer 63

lol ok lets do this.

Answer 64

\(\large \frac{ x - 2}{x^2 - x - 6} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2(x + 2)} \)

\(\large \frac{ x - 2}{(x-3)\cancel{(x+2)}} = \frac{1}{\cancel{(x+2)}(x-2)} + \frac{3}{2\cancel{(x + 2)}} \)

\(\large \frac{ x - 2}{(x-3)} = \frac{1}{(x-2)} + \frac{3}{2} \)

\(\large \frac{ x - 2}{(x-3)} = \frac{2+3(x-2)}{2(x-2)}\)

\(\large \frac{ x - 2}{(x-3)} = \frac{2+3x-6)}{2(x-2)}\)

\(\large \frac{ x - 2}{(x-3)} = \frac{3x-4)}{2(x-2)}\)

Answer 65

i just took the LCD of right side, so we can write it as single term. instead of 2 terms.
u knw LCD ?

Answer 66

least common denominator?

Answer 67

exactly ! LCD of 2, (x-2) is simply 2(x-2)

Answer 68

let me knw once u follow upto above steps. we can go ahead with rest of the problem, then

Answer 69

i follow

Answer 70

great :)

Answer 71

\(\large \frac{ x - 2}{x^2 - x - 6} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{x^2-4} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2x + 4} \)

\(\large \frac{ x - 2}{(x-3)(x+2)} = \frac{1}{(x+2)(x-2)} + \frac{3}{2(x + 2)} \)

\(\large \frac{ x - 2}{(x-3)\cancel{(x+2)}} = \frac{1}{\cancel{(x+2)}(x-2)} + \frac{3}{2\cancel{(x + 2)}} \)

\(\large \frac{ x - 2}{(x-3)} = \frac{1}{(x-2)} + \frac{3}{2} \)

\(\large \frac{ x - 2}{(x-3)} = \frac{2+3(x-2)}{2(x-2)}\)

\(\large \frac{ x - 2}{(x-3)} = \frac{2+3x-6)}{2(x-2)}\)

\(\large \frac{ x - 2}{(x-3)} = \frac{3x-4)}{2(x-2)}\)

Cross multuply

\(\large 2(x-2)^2 = (3x-4)(x-3)\)

Answer 72

how do i set that up to cross multiply?