Question

"If x>2, then x^2-x-6/x^2-4=??" (Please show your work, Thanks!)

Answer 1

what do you get when you factor the numerator?

Answer 2

Umm...hehe...Don't the two x^2's cancel each other out...? (sorry... :/ )

Answer 3

no that's a common misconception, but they don't

Answer 4

no

Answer 5

what two numbers multiply to -6 and add to -1?

Answer 6

if you're not sure, list out the possible ways to factor -6


-6 = -6*1
-6 = -3*2
-6 = -2*3
-6 = -1*6


now which pair of factors add to -1 ?

Answer 7

-2*3

Answer 8

close

Answer 9

-2+3 = 1 not -1

Answer 10

OH the other one: -3*2
but why -6 and -1?

Answer 11

I was just listing all the possible ways to multiply two whole numbers to get -6

Answer 12

yes -3+2 = -1

Answer 13

the two numbers are -3 and 2

so x^2 - x - 6 factors to (x-3)(x+2)

Answer 14

now do the same for x^2-4

and think of x^2-4 as x^2+0x-4

Answer 15

For Numerator
u can always use this formula for solving such equations:
ax^2+bx+c=0
then roots are
x=[-b+Squrt(b^2-4ac)]/2a & [-b-sqrt(b^2-4ac)]/2a

For Denominator
There is an identity
x^2-a^2=(x+a)(x-a)

Then cancel out the common terms
This should help

Answer 16

so what do you get when you factor x^2-4 or x^2+0x-4

Answer 17

So...now find a factor for -4?

Answer 18

correct list all the possible ways to multiply to -4

Answer 19

see which factorization pair adds to 0

Answer 20

Proceed @jim_thompson5910 is saying right now but after completing this sum . Remember those identities . it will be helpful for future problem solving

Answer 21

2*-2= -4
and 2+(-2)=0

Answer 22

good

Answer 23

so x^2 - 4 factors to (x+2)(x-2)

Answer 24

you now have

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

Answer 25

I'm sure you see what cancels

Answer 26

(x+2)/(x-2) cancels

Answer 27

so now left with (x-3)/(x+2)?

Answer 28

the x+2 terms cancel I think you meant to say

Answer 29

only identical terms can cancel because x/x = 1

Answer 30

right, sorry

Answer 31

so you're left with

x-3
-------
x-2

Answer 32

Kay

Answer 33

Thanks a lot @jim_thompson5910 ! Really appreciate it!

Answer 34

Now u can have a look at these :)

Answer 35

you're welcome