Factoring Trouble

hgraff

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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. :-(
There is no such number

\(\displaystyle 2x = 60 \implies \frac{1}{2} * 2x = \frac{1}{2} * 60 \implies x = 30 \implies x^2 = 900 \ne 890.\)
 
Hello, Bob Brown!

Very imaginative!
I got a different answer.


Yes it is 30, if all numbers are base -3/2

Let \(\displaystyle b\) = base, \(\displaystyle bx+y\) = two-digit number.

We have: .\(\displaystyle \begin{Bmatrix}(bx+y)^2 \;=\; 8b^2+9b & [1] \\ 2(bx+y) \;=\; 6b & [2] \end{Bmatrix}\)

From [2]: .\(\displaystyle y \:=\:3b-bx\)

Substitute into [1]: .\(\displaystyle [bx + (3b-bx)]^2 \:=\:8b^2 + 9b\)

. . . \(\displaystyle (3b)^2 \:=\:8b^2+9b \quad\Rightarrow\quad 9b^2 \:=\:8b^2 + 9b\)

. . . \(\displaystyle b^2 - 9b \:=\:0 \quad\Rightarrow\quad b(b-9) \:=\:0\)

. . . \(\displaystyle \color{red}{\rlap{/////}}{b = 0},\;b = 9\)


We are dealing with base-nine.
. . And it turns out that \(\displaystyle x = 3,\;y = 0.\)


Check:

. . \(\displaystyle \begin{Bmatrix}(30_9)^2 \;=\;8\color{purple}{9}0_9 & \Rightarrow & 27^2 \:=\:729 & \checkmark \\ 2(30_9) \;=\;60_9 & \Rightarrow & 2(27) \:=\:54 & \checkmark \end{Bmatrix}\)


Unfortunately, there is no "9" in base-nine . . . *sigh*
 

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 \(\displaystyle 2x^2 - x + 3 < 0 \)

(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.
 
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