Question

Can someone help me find the inverse of http://media.education2020.com/evresources/649-04-04-00-00_files/i0160000.jpg

Answer 1

for the inverse of a fraction, you need them to form a 1 when they multiply.

Answer 2

oh no
it is
\[g(x)=\frac{3}{x^2+2x}\] right?

Answer 3

we can do this by putting
\[x=\frac{3}{y^2+2y}\] and solving this equation for \(y\)

Answer 4

can you do that, or would you like me to walk you through it?

Answer 5

Can you walk me through it? I'm a bit confused on the steps of how to solve it.

Answer 6

yeah this one is a pain, because of the \(y^2\)
we will have to solve a quadratic equation

Answer 7

\[x=\frac{3}{y^2+2y}\]first multiply by the denominator to get
\[x(y^2+2y)=3\] or
\[xy^2+2xy-3=0\]

Answer 8

then use the quadratic formula
\[y=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]with
\[a=x, b=2x, c=-3\] don't worry that these are variables, just plug them in to get your answer

Answer 9

\[y=\frac{-2x\pm\sqrt{(2x)^2-4\times x\times (-3)}}{2x}\] is the substitution

Answer 10

we can clean it up a bit
need help with that too?

Answer 11

omg thats a bit confusing @_@

Answer 12

ikr

Answer 13

but it is just substituting in to the quadratic formula

Answer 14

lol, im still trying to process it

Answer 15

don't forget you are solving for \(y\) so when you see
\[xy^2+2xy-3=0\] think of it as
\[\color{red}xy^2+\color{blue}{2x}y\color{green}{-3}=0\] with coefficients
\[\color{red}a=\color{red}x,\color{blue}{b}=\color{blue}{2x}, \color{green}{c}=\color{green}{-3}\]

Answer 16

really no different from
\[4y^2+8y-15=0\] except the coefficients have an \(x\) in them

Answer 17

let me know when you are ready to continue, because we are still not quite done

Answer 18

LOL, I was trying to solve \[y=\frac{-2x\pm\sqrt{(2x)^2-4\times x\times (-3)}}{2x} \]
on my own.

Answer 19

good
because that is what you have to do

Answer 20

I don't understand how to do it under the square root ><

Answer 21

just do it

Answer 22

2x^2 + 12x is what i got for the one under the square root

Answer 23

oops

Answer 24

you forgot to square the 2

Answer 25

\[(2x)^2-4\times x\times (-3)=4x^2+12x\]

Answer 26

still a couple more steps

Answer 27

\[y=\frac{-2x\pm\sqrt{(2x)^2- 12x}}{2x}\]

Answer 28

\[y=\frac{-2x\pm\sqrt{(2x)^2+ 12x}}{2x}\]

Answer 29

oh whoopsss the + sign. haha

Answer 30

you have two minus signs there, so it is plus

Answer 31

not done yet

Answer 32

\[4x^2+12x=4(x^2+3x)\] and so
\[\sqrt{4x^2+12x}=\sqrt{4(x^2+3x)}=\sqrt{4}\sqrt{x^2+3x}=2\sqrt{x^2+3x}\]

Answer 33

not done yet

Answer 34

woeee you just lost me D:

Answer 35

\[y=\frac{-2x\pm\sqrt{(2x)^2-4\times x\times (-3)}}{2x}\]
\[=y=\frac{-2x\pm2\sqrt{x^2+3x}}{2x}\]\[=\frac{2(-x\pm\sqrt{x^2+3x})}{2x}\]
\[=\frac{-x\pm\sqrt{x^2+3x}}{x}\]

Answer 36

ok let me try to unlose you

Answer 37

in simplest radical form
\[\sqrt{40}=2\sqrt{10}\] because
\[40=4\times 10\] and so
\[\sqrt{40}=\sqrt{4}\sqrt{10}=2\sqrt{10}\]

Answer 38

i did the same thing in the numerator, factored out the common factor of 4 inside the square root, and then pulled it outside of the radical

Answer 39

then you have a common factor of 2 in both terms of the numerator, which cancels with the 2 in the denominator

Answer 40

oh i see where you got that now.

Answer 41

k good
because finally we are done

Answer 42

although you see that the inverse is NOT a function, because it has that \(\pm\) in it

Answer 43

im on \[=\frac{2(-x\pm\sqrt{x^2+3x})}{2x} \]

Answer 44

good
now cancel the common factor of 2 top and bottom

Answer 45

okay

Answer 46

now i have a question
what is "impact academy"?

Answer 47

It's an online school :P

Answer 48

but waitt!
do we also cancel the x?

Answer 49

oooh

Answer 50

no you cannot cancel the \(x\) because \(x\) is not a factor of the numerator

Answer 51

that would be like cancelling here
\[\frac{2+5}{2}=\frac{\cancel{2}+5}{\cancel{2}}+5\]

Answer 52

Aaaah, is there another way this problem can be written? My quiz doesnt have this one as its option D:

Answer 53

what are the options?

Answer 54

a. http://media.education2020.com/evresources/649-04-04-00-00_files/i0160002.jpg
b. http://media.education2020.com/evresources/649-04-04-00-00_files/i0160004.jpg
c. http://media.education2020.com/evresources/649-04-04-00-00_files/i0160003.jpg
d. http://media.education2020.com/evresources/649-04-04-00-00_files/i0160005.jpg

Answer 55

jeez they are goofy
let me see if i can figure out which one it is

Answer 56

\[\frac{x}{x}=1\] is right, so the one out front is good

Answer 57

T___T

Answer 58

ok got it

Answer 59

I don't know how they got those answers :o

Answer 60

\[\frac{\sqrt{x^3+3x}}{x}\]
\[=\sqrt{\frac{x^2+3x}{x^2}}\]
\[=\sqrt{1+\frac{3}{x}}\]

Answer 61

go with A

Answer 62

no no scratch that!!

Answer 63

:O

Answer 64

I have 5 minutes left ;(

Answer 65

i get
\[-1\pm\sqrt{1+\frac{x}{3}}\]

Answer 66

but i checked the computer and it gives a 1 instead
i like the -1

Answer 67

I guess I'll go with B :O

Answer 68

@wio you got a suggestion?

Answer 69

I have one more question but i only have 2 minutes lol. I think im just gonna guess the last one

Answer 70

ok good luck

Answer 71

i would go with A for this one

Answer 72

What is the issue?

Answer 73

i get
\[-1\pm\sqrt{\frac{x}{3}+1}\] but i put it in wolfram and it gave me a 1 out front instead of -1

Answer 74

i can't figure it

Answer 75

Multiply it by \(-1\).

Answer 76

I submitted my quiz. @satellite73 you were right :P

Answer 77

It is a bit strange.

Answer 78

can't just multiply something by -1!!

Answer 79

it is -1 though. idk

Answer 80

yeah i guess wolfram slept on this one

Answer 81

Thanks a lot though :D I got a 90.