a)101.22= 6/1+y + 106/(1+y)^2

b) 101.22= 112+6y/(1+y)^2

c) 101.22y^2 +202.44y+101.22= 112+6y

and i was wondering how they got there from a to c

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- Feb 18th 2011, 03:59 AMmatlondonstrange quadratic
a)101.22= 6/1+y + 106/(1+y)^2

b) 101.22= 112+6y/(1+y)^2

c) 101.22y^2 +202.44y+101.22= 112+6y

and i was wondering how they got there from a to c - Feb 18th 2011, 04:04 AMtopsquark
Could you put the appropriate parenthesis in here? I can make out what b) is supposed to be, but only because I see how to go from b) to c). (Multiply both sides of the equation by (1 + y)^2 and expand.) But I can't make any sense out of how to go from a) to b). There are too many ambiguities in a).

b) 101.22= (112+6y)/(1+y)^2

c) 101.22y^2 +202.44y+101.22= 112+6y

-Dan

PS Oh I get a) now. It's supposed to be

a) 101.22= 6/(1+y) + 106/(1+y)^2

Add the two fractions:

etc. - Feb 18th 2011, 04:07 AMe^(i*pi)
I suspect a) is meant to read

- Feb 18th 2011, 04:12 AMSwlabr
Assuming e^(i*pi) is correct, to get your answer simply multiply through by (1+y)^2, expand, and then you'll have something more familiar. (If you're feeling daring, there is no need to expand - you can set x=1+y, multiply through by x^2 and solve for x. Then y=x-1...)

- Feb 18th 2011, 04:20 AMmatlondon
- Feb 18th 2011, 04:21 AMmatlondon
- Feb 18th 2011, 04:22 AMe^(i*pi)
You need brackets on your first term on the RHS. At the moment it reads whereas I suspect you mean

It's the same principle as Swlabr stated in post 4. However, this time, you need to multiply through by (1+y)^3 and then expand

Quote:

how would i do that "(If you're feeling daring, there is no need to expand - you can set x=1+y, multiply through by x^2 and solve for x. Then y=x-1...)"

or if you can't show me could you direct me to a site that does.

when you multiply by u^2 it gives

Solve for u (I suggest the quadratic formula) and then back substitute to find y. (Remember that -1 is not in the domain)

Personally I think it's easier than expanding! - Feb 18th 2011, 04:27 AMArchie Meade
- Feb 18th 2011, 04:49 AMmatlondon
- Feb 19th 2011, 04:07 AMHallsofIvy
Or, let so that [tex]101.22= \frac{6}{1+y}+ \frac{106}{(1+y)^2}[\math] becomes

Solve that for x, then

For [tex]101.22= \frac{6}{1+y}+ \frac{6}{(1+y)^2}+ \frac{106}{(1+ y)^3}[tex]

the same substitution give

Of course, cubics are considerably harder to solve than quadratics.