# Difference of two numbers

• July 27th 2006, 06:02 AM
pashah
Difference of two numbers
Hello I am having some difficulty with this problem.

Two positive numbers differ by 11, and their square roots differ by 1. Find
the numbers.

Difference between the numbers is 11.

This means (x-y)=11
substituting in our main equation, we get:
11*(x+y)=1
(x+y)=0.5

Now we arrive at two simple linear equations:
(x+y)=0.5
(x-y)=11

solve for 'x' and 'y' from the above eqns to get:
x=5.75, y=-5.25
• July 27th 2006, 06:13 AM
CaptainBlack
Quote:

Originally Posted by pashah
Hello I am having some difficulty with this problem.

Two positive numbers differ by 11, and their square roots differ by 1. Find
the numbers.

Difference between the numbers is 11.

This means (x-y)=11
substituting in our main equation, we get:
11*(x+y)=1
(x+y)=0.5

Now we arrive at two simple linear equations:
(x+y)=0.5
(x-y)=11

solve for 'x' and 'y' from the above eqns to get:
x=5.75, y=-5.25

$
x-y=11
$

$
\sqrt{x}-\sqrt{y}=1
$

Trial and error indicates 36 and 25 are suitable numbers.

RonL
• July 27th 2006, 06:21 AM
pashah
Thanks Capn
• July 27th 2006, 01:03 PM
ThePerfectHacker
Quote:

Originally Posted by pashah
Hello I am having some difficulty with this problem.

Two positive numbers differ by 11, and their square roots differ by 1. Find
the numbers.

Difference between the numbers is 11.

This means (x-y)=11
substituting in our main equation, we get:
11*(x+y)=1
(x+y)=0.5

Now we arrive at two simple linear equations:
(x+y)=0.5
(x-y)=11

solve for 'x' and 'y' from the above eqns to get:
x=5.75, y=-5.25

A more elegant solution.

You have,
$x-y=11$
Use diffrence of two squares,
$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=11$
But,
$\sqrt{x}-\sqrt{y}=1$
Thus,
$\sqrt{x}+\sqrt{y}=11$
Therefore, the two equation you have are,
$\sqrt{x}+\sqrt{y}=11$
$\sqrt{x}-\sqrt{y}=1$
$2\sqrt{x}=11\rightarrow x=36$
$2\sqrt{y}=10\rightarrow x=25$
• July 27th 2006, 06:25 PM
pashah
You Reign
Thank you for the expert help. One thing you did forget to mention about the four individuals. They were all megalomaniacs who were hell bent on conquering the known world. Unlike yourself I don't see where they were interested in helping others overcome difficulties. Thanks Again.
• July 27th 2006, 06:38 PM
ThePerfectHacker
Quote:

Originally Posted by pashah
Thank you for the expert help. One thing you did forget to mention about the four individuals. They were all megalomaniacs who were hell bent on conquering the known world. Unlike yourself I don't see where they were interested in helping others overcome difficulties. Thanks Again.

I would place Hannibal of Carthage on the list since he is my favorite (ever I seen Hannibal vs. Rome on Histroy Channel Special) but he never did try to take over the world.
• July 27th 2006, 08:43 PM
pashah
Hannibal the Barca
That's a strange coincidence since I am also fond of Hannibal. His father Hamilcar was quite an impressive leader as well. Although, I must say they seem to have commited some horrible atrocities in their conquests. Hannibal was known to slaughter entire companies of his own men for fear that they would fall prey to the Roman legions of Scipio. I suspect his biggest mistake was limiting his tactics. He was redundant and consequently an observant Scipio would later adopt his tactics and use those very same tactics to defeat him in a decisive battle.

Did you know that Hannibal was originally from an ancient tribe of Spanish descent?
• July 27th 2006, 09:05 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I would place Hannibal of Carthage on the list since he is my favorite (ever I seen Hannibal vs. Rome on Histroy Channel Special) but he never did try to take over the world.

Won spectacular victories in battle, lost war due to inability to
overcome political constraints.

Compare with oneone like W S Churchill - dreadfull when interfereing
with the running of campaigns but understood the (geo-) political
side better than the enemies of the UK - result: the UK on winning side
in one of the most important conflicts in its history.

RonL
• July 29th 2006, 07:40 PM
Soroban
Hello, pashah!

TPHacker's solution is elegant.
It still can be solve by "normal" methods.

Quote:

Two positive numbers differ by 11, and their square roots differ by 1.
Find the numbers.

We have: . $\begin{array}{cc}(1)\;\;x \:- \:y\;=\;\;11 \\ (2)\;\sqrt{x} - \sqrt{y}\:=\:1\end{array}$

From (2), we have: . $\sqrt{y} \,= \,\sqrt{x}-1\quad\Rightarrow\quad y \,= \,(\sqrt{x} - 1)^2$

Substitute into (1): . $x - (\sqrt{x} - 1)^2\:=\:11\quad\Rightarrow\quad x - (x - 2\sqrt{x} + 1) \:= \:11$

Hence: . $2\sqrt{x} - 1 \:=\:11\quad\Rightarrow\quad 2\sqrt{x}\,=\,$ $12\quad\Rightarrow\quad \sqrt{x}\,=\,6\quad\Rightarrow\quad x = 36$

Therefore: . $\boxed{x = 36,\;y = 25}$