All I need to know is what number, when squared, equals 890, and when doubled, equals 60. No, it's not 30. Any ideas?
Perhaps there's no solution. IDK.
All I need to know is what number, when squared, equals 890, and when doubled, equals 60. No, it's not 30. Any ideas?
Perhaps there's no solution. IDK.
Hello, Bob Brown!
Very imaginative!
I got a different answer.
Let [tex]b[/tex] = base, [tex]bx+y[/tex] = two-digit number.
We have: .[tex]\begin{Bmatrix}(bx+y)^2 \;=\; 8b^2+9b & [1] \\ 2(bx+y) \;=\; 6b & [2] \end{Bmatrix}[/tex]
From [2]: .[tex]y \:=\:3b-bx[/tex]
Substitute into [1]: .[tex][bx + (3b-bx)]^2 \:=\:8b^2 + 9b[/tex]
. . . [tex](3b)^2 \:=\:8b^2+9b \quad\Rightarrow\quad 9b^2 \:=\:8b^2 + 9b[/tex]
. . . [tex]b^2 - 9b \:=\:0 \quad\Rightarrow\quad b(b-9) \:=\:0[/tex]
. . . [tex]\color{red}{\rlap{/////}}{b = 0},\;b = 9[/tex]
We are dealing with base-nine.
. . And it turns out that [tex]x = 3,\;y = 0.[/tex]
Check:
. . [tex]\begin{Bmatrix}(30_9)^2 \;=\;8\color{purple}{9}0_9 & \Rightarrow & 27^2 \:=\:729 & \checkmark \\ 2(30_9) \;=\;60_9 & \Rightarrow & 2(27) \:=\:54 & \checkmark \end{Bmatrix}[/tex]
Unfortunately, there is no "9" in base-nine . . . *sigh*
So it is just possible that Bob Brown was right all along?
Instead of "possible," it is definite.
http://en.wikipedia.org/wiki/Negative_base
State it with conviction.
As a tangential example,
one would state with conviction that [tex]2x^2 - x + 3 < 0 [/tex]
(or similar expression) would not be presented as, say,
2xx - x + 3 < 0.
--------------------------------------------
Edit: "Definitively missing the joke."
Or I "got the joke" and deliberately was dismissive of it.
Last edited by lookagain; 02-20-2013 at 05:42 PM.
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