# Homework Help: Solve quadratic in non-10 base

1. Oct 13, 2012

### zaper

1. The problem statement, all variables and given/known data
I need to solve 0 = 5x2 - 50x + 125 with solutions x = 5, x = 8.

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

x2 - 13x + 40 = 5x2 - 50x + 125

3. The attempt at a solution

I tried to simply compare coefficients here but I saw that for x2 this would give me 1b = 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

1b = 110 -> 1 = 1
-13b = -1010 -> 1 * b + 3 = 10
40b = 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.

2. Oct 13, 2012

### rcgldr

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)?

Last edited: Oct 13, 2012
3. Oct 13, 2012

### 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 = 5b

4. Oct 13, 2012

### rcgldr

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

5. Oct 13, 2012

### gabbagabbahey

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.

6. Oct 13, 2012

### 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

7. Oct 13, 2012

### rcgldr

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