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- anonymous

i have 2 equations
(4x3+24x2+7x+42)/(x+6)and(2x2+6x+a)
the first formula has an x<-6 and the 2nd has x

Mathematics
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- anonymous

i have 2 equations
(4x3+24x2+7x+42)/(x+6)and(2x2+6x+a)
the first formula has an x<-6 and the 2nd has x

Mathematics
- chestercat

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- anonymous

factor and cancel to get
\[4x^2+7\]

- anonymous

replace x by 6 and get
\[4\times 6^2+7=189\]

- anonymous

so my algebra is wrong but my method is right?

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- anonymous

what you then have to do, now that you know the limit of the first expression is equal to 189, is replace x by 6 in the second expression, set it also equal to 189 and solve for "a"

- anonymous

there is only 1 a though

- anonymous

and that is in the 2nd equation

- anonymous

right. the second expression is
\[2x^2+6x+a\] and you need to to match up to the first expression when x =6

- anonymous

you know the limit of the first expression at 6 is 186 so write
\[2\times 6^2+6\times 6+a=189\] and solve for "a"

- anonymous

sorry i meant "189"

- anonymous

oh ok that makes sense

- anonymous

i get
\[108+a=189\] and therefore
\[a=81\] but don't take my word for it

- anonymous

it says 81 is in incorrect

- anonymous

x had to equal -6 not 6

- anonymous

x=-7, a=147
x=-8, a=183 and so on

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