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

[lcliff]

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???

 

[soroban]

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!

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

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

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

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

\(\displaystyle \;\;10\,-\,y^2\:=\:2y\)

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

then by quad formula I got; \(\displaystyle \,y\:\approx\:-4.31\) and \(\displaystyle 2.31\)

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

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


Check: \(\displaystyle \,x\,-\,y\:=\:4.31\,-\,2.31 \:=\:2\)

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

Nice work!