word problem: Find two positive real numbers that....

lcliff

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May 31, 2006
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Missing numbers. Find two possitive real numbers that differ by 2 and have a product of 10.

This is how I started but I really do not think it is right!

. . .x - y = 2, xy = 10
. . .10/y - y = 2, x = 10/y
. . .10/y - y^2/y = 2
. . .(10 - y^2)/y = 2
. . .10 - y^2 = 2y
. . .-y^2 - 2y + 10 = 0

Then, by the Quadratic Formula, I got:

. . .-4.31 and 2.31

. . .2.31 + 4.31 =(apx) 2
. . .2.31*4.31 =(apx) 10

Is this right???
 
Re: need help with this problem

Hello, lcliff!

You did great!


Find two positive real numbers that differ by 2 and have a product of 10.

This is how I started but I really do not think it is right!

    x  y=  2      xy=10        x=10y\displaystyle \;\;x\,-\;y\:=\;2\;\;\;xy\:=\:10\;\;\Rightarrow\;\;x\,=\,\frac{10}{y}

    10yy=2\displaystyle \;\;\frac{10}{y}\,-\,y\:=\:2

    10yy2y=2\displaystyle \;\;\frac{10}{y}\,-\,\frac{y^2}{y}\:=\:2

    10y2y=2\displaystyle \;\;\frac{10\,-\,y^2}{y}\:=\:2

    10y2=2y\displaystyle \;\;10\,-\,y^2\:=\:2y

\(\displaystyle -y^2\,-\,2y\.+\,10\:=\:0\)

then by quad formula I got; y4.31\displaystyle \,y\:\approx\:-4.31 and 2.31\displaystyle 2.31

Since x\displaystyle x and y\displaystyle y are positive numbers: y=2.31\displaystyle y\:=\:2.31

Then: x=y+2=2.31+2=4.31\displaystyle \,x\:=\:y\,+\,2\:=\:2.31\,+\,2\:=\:4.31


Check: xy=4.312.31=2\displaystyle \,x\,-\,y\:=\:4.31\,-\,2.31 \:=\:2

. . . . . . . .xy  =  (4.31)(2.31)=9.93310\displaystyle xy \;=\;(4.31)(2.31) \:=\: 9.933\:\approx\:10


Nice work!

 
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