x/(x^2 - 2) = 2/(3x + 4) solve the proportion

we can "cross multiply" and then collect the terms to one side and solve a quadratic i believe

uh ok can you do that for me I'm taking a pretest so i haven't learned this yet

the pretest is the place TO learn it :)

please i need a good grade on this

\[\frac{a}{b}=\frac{c}{d}\to \ ad=bc\]

thats the essense of a cross multiply

um ok i'll try that so x(3x+4)= 2(x^2 - 2)

x(3x+4) = 2(x^2 - 2) i agree so far; now lets distribute thru on each side ...

\[x(a+b)\to \ ax+bx\] is distribution

ok so 4x + 4x = 2x^2 - 4 ???

almost; that 3xx kinda snuck away from you :) 3xx+4x = 2x^2 - 4 3x^2 +4x = 2x^2 - 4 now just move all the terms to one side using the appropriate adds and subtracts

3x^2 +4x = 2x^2 - 4 -2x^2 +4 -2x^2+4 ------------------------ x^2 +4x +4 = 0

now i do something with like quadratic equation right?

yes, and depending on how familiar you are with them this can be easy or tricky ...

it factors nicely tho if your comfy with that

yeah i don't really remember how to do it

yeah, it takes practice :) that end is a 4; we need the factors of 4: 1*4 2*2 the middle number is a 4; we need a set of factors from that first part that ADD up to 4 1+4 = 5 ; not it 2+2 = 4 ; that one will do just fine split the quadratic into its component parts and fill in our numbers (x+2) (x+2) = 0 is then what we solve

when x=-2 we get: (-2+2) (-2+2) = 0 0(0) = 0 0=0 so when x=-2 is the solution to all our problems .... in this case :)

oh yesss now i remember so x= 2 and x = -2 ???

or is it just x= -2

just x=-2 ... thats the only thing that will fit

if we try x=2 (2+2)(2+2)=0 4(4) = 0 16=0 just doesnt fit

okay thanks can u answer my other question. its the last question i asked.

i can give it a shot

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