Question

Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3?

Answer 1

4/x + 4/(x^2) - 9 = 3/(x - 3)?

Did I do this right?

Answer 2

\[\frac{ 4 }{ x }+\frac{ 4 }{ x^2-9 }=\frac{ 3 }{ x-3 }\]

Answer 3

how else can you write
\[x^2-9\]

Answer 4

does the format ring bell?

Answer 5

(x - 3i)(x + 3i), where i = √(-1), right?

Answer 6

no

Answer 7

where did i come from

Answer 8

\[i= \sqrt -1\]

Answer 9

it is just (x+3)(x-3)

Answer 10

Oh

Answer 11

yes

Answer 12

now base is not equal
do you know how to add numbers with unequal bases?

Answer 13

no

Answer 14

\[\frac{ a }{ b }+ \frac{ c }{ d } = \frac{ a*d+b*c }{ b*d }\]

Answer 15

put the values tell me what you got

Answer 16

\[\frac{ ((4x+12)(4x-12))+4x }{ (x^2+3x)(x^2-3x) }\]

Answer 17

~_~ why did you multiply 4 to both the brackets

Answer 18

I shouldn't have done that?

Answer 19

no

Answer 20

So only multiply 4 to the first bracket?

Answer 21

yeah
do this
multiply by x^2-9

Answer 22

keep the base value same

Answer 23

which is x(x+3)(x-3)

Answer 24

so, do I multiply 4 and (x^2-9)?

Answer 25

yes

Answer 26

i will save your work
wait

Answer 27

Okay

Answer 28

\[\frac{ 4x^2+4x-12 }{ x(x+3)(x-3) } =\frac{ 3 }{ x-3 }\]

Answer 29

now take 4 common
you will get 4(x^2+x-12)

Answer 30

factorize the term in bracket then you will get 4(x-1)(x+3)

Answer 31

And now I factor it to \[4(x-3)(x+4)\]

Answer 32

\[\frac{ 4(x-1)(x+3) }{ x(x+3)(x-3) } =\frac{ x }{ (x-3) }\]

Answer 33

hold on

Answer 34

So then, it's \[\frac{ 4(x-3)(x+4) }{ x(x+3)(x-3) }=\frac{ x }{ (x-3) }\]?

Answer 35

4*9 = 36 ~_~
sorry i got the equation wrong

Answer 36

okay, so what would the correct equation be?

Answer 37

\[4x^2+4x-36\]

Answer 38

okay

Answer 39

so take out 4
you will get x^2+x-9
now solve

Answer 40

factorized, it is (x-3)(x+4)

Answer 41

You don't want to expand anything.
Keeping things factored makes it easy to see roots.

Answer 42

(x-3)(x+4) ? = x^2+x+12

Answer 43

-12 **

Answer 44

Nevermind, you will have to expand at some point.

Answer 45

i guess

Answer 46

there are no real roots for x^2+x-9

Answer 47

So x²+x-9 is not factorable under the reals, however it shouldbr factorable under the complex numbers. If i'm correct, then

(x^2+x-9)=(x - 3i)(x + 3i), where i = √(-1) right?

Answer 48

*should be

Answer 49

LOL dont go for complex numbers
this is not related to complex numbers

Answer 50

okay

Answer 51

So complex numbers would just complicate it more

Answer 52

here is the solved 1
when you cancel equal factors x-3
you will get
\[4x^2+4x-36=3x(x+3)\]

Answer 53

last equation will be
\[x^2-5x-36=0\]

Answer 54

which is
\[(x-9)(x+4)=0\]

Answer 55

so your answer is
x= 9 or x= -4

Answer 56

Okay, Thanks!

Answer 57

you are welcome
@wio thanks
i would hv got lost factorizing this :D