strange quadratic

Printable View

February 18th 2011, 03:59 AM
matlondon

strange 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
February 18th 2011, 04:04 AM
topsquark

Quote:

Originally Posted by matlondon

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

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:
$\displaystyle \frac{6}{1+y} + \frac{106}{(1+y)^2}$

$\displaystyle = \frac{6(1 + y)}{(1 + y)^2} + \frac{106}{(1+y)^2}$

$\displaystyle \frac{6(1 + y) + 106}{(1 + y)^2}$

etc.
February 18th 2011, 04:07 AM
e^(i*pi)

I suspect a) is meant to read $101.22 = \dfrac{6}{1+y} + \dfrac{106}{(1+y)^2}$
February 18th 2011, 04:12 AM
Swlabr

Quote:

Originally Posted by matlondon

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

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...)
February 18th 2011, 04:20 AM
matlondon

$101.22 = \dfrac{6}{1+y} + \dfrac{6}{(1+y)^2} + \dfrac{106}{(1+y)^3}$
February 18th 2011, 04:21 AM
matlondon

Quote:

Originally Posted by Swlabr

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...)

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.
February 18th 2011, 04:22 AM
e^(i*pi)

You need brackets on your first term on the RHS. At the moment it reads $6 + y$ whereas I suspect you mean $\dfrac{6}{1+y}$

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.

$u = 1+y \implies y = u-1$

$101.22 = \dfrac{6}{1+y} + \dfrac{106}{(1+y)^2} \implies 101.22 = \dfrac{6}{u} + \dfrac{106}{u^2}$

when you multiply by u^2 it gives $101.22u^2 = 6u + 106 \implies 101.22u^2-6u-106 = 0$

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!
February 18th 2011, 04:27 AM
Archie Meade

Quote:

Originally Posted by matlondon

thanks guys so what if it was 101.22= 6/1+y + 6/(1+y)^2+106/(1+y)^3

$101.22=\displaystyle\frac{6}{1+y}+\frac{6}{(1+y)^2 }+\frac{106}{(1+y)^3}$

$\displaystyle\ 101.22=\frac{6(1+y)^2}{(1+y)^3}+\frac{6(1+y)}{(1+y )^3}+\frac{106}{(1+y)^3}$

Now we are adding the same fractions

$\displaystyle\ 101.22=\frac{6(1+y)^2+6(1+y)+106}{(1+y)^3}$

$101.22(1+y)^3=6(1+y)^2+6(1+y)+106$

and multiply out...
February 18th 2011, 04:49 AM
matlondon

Quote:

Originally Posted by e^(i*pi)

You need brackets on your first term on the RHS. At the moment it reads $6 + y$ whereas I suspect you mean $\dfrac{6}{1+y}$

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

$u = 1+y \implies y = u-1$

$101.22 = \dfrac{6}{1+y} + \dfrac{106}{(1+y)^2} \implies 101.22 = \dfrac{6}{u} + \dfrac{106}{u^2}$

when you multiply by u^2 it gives $101.22u^2 = 6u + 106 \implies 101.22u^2-6u-106 = 0$

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!

U=1+y/2
[MATH\]dfrac{6}{u} + dfrac{6}{u^2}\+dfrac{6}{u^3}dfrac{106}{u^4}[/tex]
how would i do that , is there any sites that can help me with it
February 19th 2011, 04:07 AM
HallsofIvy

Or, let $x= \frac{1}{y+ 1}$ so that [tex]101.22= \frac{6}{1+y}+ \frac{106}{(1+y)^2}[\math] becomes
$101.22= 6x+ 106x^2$

Solve that for x, then $y= \frac{1}{x}- 1$

For [tex]101.22= \frac{6}{1+y}+ \frac{6}{(1+y)^2}+ \frac{106}{(1+ y)^3}[tex]
the same substitution give
$101.22= 6x+ 6x^2+ 106x^3$
Of course, cubics are considerably harder to solve than quadratics.