# Finding the base of numbers from a given equation

1. Apr 3, 2010

### kliker

x^2 - 11x + 22 = 0

the roots are x= 3 and x = 6

what's the base of the numbers?

ok what I've done is I found the roots in decimal system which are

x1 = (11+sqrt(33))/2 and x2 = (11-sqrt(33)/2

then I said x1 = 3 and x2 = 6 so we have

3 = (11+sqrt(33)/2 <=> 6 = 11+sqrt(33) <=> -5 = sqrt(33) <=> 25 = 33 <=> 2b + 5 = 3b + 3 <=> b = 2

im doing the same for the second root

6 = (11-sqrt(33)/2 <=> 12 = 11 -sqrt(33) <=> 1 = -sqrt(33) <=> 1 = 33 <=> 1 = 3b + 3<=>
3b = -2 <=> b = -2/3

i know that my method is wrong, the correct answer is 8 but that's what I thought would solve this problem, it's the first time that I see this kind of problem, so if anyone could guide me I would apreciate it

2. Apr 3, 2010

### Petek

Do you know the formulas that express the coefficients (-11 and 22) of the polynomial in terms of its roots (3 and 6)?

Petek

3. Apr 3, 2010

### kliker

i'm not sure, do you mean something like this:

22 = 6 <=> 2b + 2 = 6 <=> b=2

sorry im not that good in english :(

Last edited: Apr 3, 2010
4. Apr 3, 2010

### Petek

Suppose that the polynomial $x^2 + ax + b$ has roots r and s. Then we can write $x^2 + ax +b = (x - r)(x - s)$. Expand the expression on the right side of the equation. You then should be able to get formulas for a and b in terms of r and s. Do you see how this helps to solve your original question?

Petek

5. Apr 3, 2010

### ramsey2879

You pick the hard way to do it but it can work if you do it right

x^2 + bx + c if The base is B then B+1 = -b and 2B+2 = c

so 3 = (-b -sqrt(b^2-4ac))/2 = (B+1 - sqrt(B^2-6B -7))/2

This is very difficult to solve but if you set B = 8 you can see it works!

Try Petek's suggestion. It is a lot easier Just remember b = -B-1 and c = 2B + 2

Last edited: Apr 3, 2010
6. Apr 4, 2010