# Finding unknown base from equation

• Jan 28th 2011, 08:36 PM
xEnOn
Finding unknown base from equation
I am given this equation of an unknown base:
$x^2 - 10x +25$
It has a solution of x=5 and x=8 which are likely to be in our base 10.

I tried to find out the unknown base by comparing the co-efficients of the equations by doing so:
$(x-5)(x-8) = x^2 - 13x + 40$ <--in base 10
$x^2 - 10x +25$ <--in unknown base

Then I try to solve for the unknown base:
1*b = 13, b=13
but...
5*1 + 2*b = 40
2b = 35
b=17.5

How can the b in the equations be different? If this is the case, what is the actual base for the equation?
thanks!
• Jan 28th 2011, 11:09 PM
CaptainBlack
Quote:

Originally Posted by xEnOn
I am given this equation of an unknown base:
$x^2 - 10x +25$
It has a solution of x=5 and x=8 which are likely to be in our base 10.

I tried to find out the unknown base by comparing the co-efficients of the equations by doing so:
$(x-5)(x-8) = x^2 - 13x + 40$ <--in base 10
$x^2 - 10x +25$ <--in unknown base

Then I try to solve for the unknown base:
1*b = 13, b=13
but...
5*1 + 2*b = 40
2b = 35
b=17.5

How can the b in the equations be different? If this is the case, what is the actual base for the equation?
thanks!

The equations you bet are inconsistent, which is what I find. So there is no solution to the problem as you post it.

Check the exact wording of the question, and post that.

Also this is not a number theory question, so I am moving the thread to the algebra section.

CB
• Jan 29th 2011, 05:08 AM
xEnOn
Thanks for moving to the correct section.

The equation from the question was: $5x^2 -50x +125$ in an unknown base with x=5 and x=8.
The question was to find the unknown base of the question. I divided by 5 on the coefficients before trying to find out the base.
• Jan 29th 2011, 11:16 AM
CaptainBlack
Quote:

Originally Posted by xEnOn
Thanks for moving to the correct section.

The equation from the question was: $5x^2 -50x +125$ in an unknown base with x=5 and x=8.
The question was to find the unknown base of the question. I divided by 5 on the coefficients before trying to find out the base.

Now use the same method as before, so in base 10:

$5(x-5)(x-8)=5x^2-(5b)x+(b^2+2b+5)$

and this time it works.

CB
• Jan 29th 2011, 06:55 PM
xEnOn
thanks CaptainBlack.
But what is the reason for adding a b this way? And why wouldn't my previous method work?

I calculated the equation and for (5b)x, b=13 but the b^2 + 2b +5, b=13.892 or -13.892. So is 13 considered the base enough though there is 0.892 more?
• Jan 30th 2011, 06:29 AM
xEnOn
ok i figured out how the 'b's come into the equation. but still, I am getting b=13 from 5b and b=13.892 or b=-13.892 from $b^2 + 2b +5$. It is weird that the 2 don't give similar results. then how can I determine its base?
• Jan 30th 2011, 08:50 AM
CaptainBlack
Quote:

Originally Posted by xEnOn
ok i figured out how the 'b's come into the equation. but still, I am getting b=13 from 5b and b=13.892 or b=-13.892 from $b^2 + 2b +5$. It is weird that the 2 don't give similar results. then how can I determine its base?

Your arithmetic is wrong, $b=13$ is a root of $200=b^2+2b+5$

CB
• Jan 31st 2011, 10:36 PM
xEnOn
oh yea! you are right about that. I had some calculation mistakes earlier..
thank you so much! :)