How Do You Solve Quadratic Equations in Non-10 Base Systems?

[zaper]

Homework Statement

I need to solve 0 = 5x² - 50x + 125 with solutions x = 5, x = 8.

Homework Equations

Referencing this thread I arrived at the equation (x - 5)(x - 8) = 5x² - 50x + 125. Expanding

x² - 13x + 40 = 5x² - 50x + 125

The Attempt at a Solution

I tried to simply compare coefficients here but I saw that for x² this would give me 1_b = 5₁₀ which I don't think is possible. So what I did instead was factor out a 5 giving

x² - 13x + 40 = 5(x² - 10x + 25) which I set up the system

1_b = 1₁₀ -> 1 = 1
-13_b = -10₁₀ -> 1 * b + 3 = 10
40_b = 25₁₀ -> 4 * b + 0 = 25

The first equation is not helpful at all, the second givers b = 7 which is not possible with x = 8 and the third gives b = 6.25 which is also not possible.

I'm really stumped here. Any help would be awesome.

 

[rcgldr]

So what I did instead was factor out a 5 giving x² - 13x + 40.

You factored out the 5 in base 10, at this point the best you can do by factoring out the 5 is x² - (50/5)x + (125/5). Also since x = 8 is a solution, then the base must be greater than 8. Also what is (50/5) in any base (any base greater than 5)?

 

[zaper]

Wow, can't believe I got those sides mixed around like that...

Ok, so then if I can't factor out 5, how would you recommend I go forward? I'm getting stuck at the 1₁₀ = 5_b

 

[rcgldr]

Ok, so then if I can't factor out 5, how would you recommend I go forward?

You can re-write the original equation as:

0 = 5 x² + (5 b + 0) x + 1 b² + 2 b + 5.

Where b is the base. You can factor out the 5 from this equation by dividing both sides by 5, but the last term will have fractions.

I'm getting stuck at the 1₁₀ = 5_b

I don't understand what you mean by this. 1 in any base = 1 in any other base.

 

[gabbagabbahey]

Homework Statement

I need to solve 0 = 5x² - 50x + 125 with solutions x = 5, x = 8.

Just to be clear (since you haven't specified this in your post), is this the quadratic equation in base 10 or your unknown base "b"? Likewise, are the solutions x = 5 & x = 8 in base 10 or base "b"?

Assuming that these are all in base "b", then you should have

\alpha_{(b)}(x_{(b)}-5_{(b)})(x_{(b)} - 8_{(b)}) = 5_{(b)}x_{(b)}^2 - 50_{(b)} x_{(b)} +125_{(b)}

That is, you should not be comparing a base "b" equation to a base 10 equation.

 

[zaper]

Ok, I just subbed in x = 5 and x = 8 and set them equal to each other. I got b = 13 which seems right to me

 

[rcgldr]

Ok, I just subbed in x = 5 and x = 8 and set them equal to each other. I got b = 13 which seems right to me

That's the correct answer. Hope we were able to help.