How to solve quadratic equation in base 13?

Question

I have to use only operations in the base 13 to solve this equation, but I don't know how to do it. I came up with the idea to convert everything in the base 10, but I need to do it in a base 13. A equation is: $$x^2 - 176_{13}\times x + 8089_{13} = 0$$

Answer 1

Numbers are numbers. Base $10$ and Base $13$ are just notations to express these numbers.

If $x^2 - 176_{13}\times x + 8089_{13} = 0$ then by quadratic formula

$x = \frac {176_{13} \pm \sqrt{176_{13}^2 - 4\times8089_{13} }}2$

So the hard part is the arithmetic.

$4\times 8089_{13} = $

$4(8* 13^3 + 8*13 + 9) = 32*13^3 + 32*13 + 36$ which if we wanted to we could reduce to $(26+6)*13^3+(26+6)*13 + (26+10)=$

$2*13^4 + 6*13^3 + 2*13^2 + (6+2)*13 + 10 = 2628A_{13}$.

And $176_{13}^2 = (13^2 + 7*13 + 6)^2 =$

$ 13^4 + (7+7)*13^3 + (6+7*7 + 6)*13^2 + (6*7+6*7)*13 + (6*6)=$

$ 13^4 + 14*13^3 + 61*13^2 + 84*13 + 36 =$

$ 13^4 + (13 +1)*13^3 + (52+9)*13^2 + (78+6)*13 + (26 + 10)=$

$ 2*13^4 + (1+4)*13^3 + (9+6)*13^2 + (6+2)*13 + 10 =$

$2*13^4 + 5*13^3 + (13+2)*13^2 + 8*13 + 10=$

$2*13^4 + 6*13^3 + 2*13^2 + 8*13 + 10= 2628A_{13}$

And so $176_{13}^2 - 4\times8089_{13}=2628A_{13}- 2628A_{13}=0$.

So $x = \frac {176_{13}}2 = \frac {13^2 + 7*13 + 6}2$

$ = \frac {(12+1)*13 + (6+1)*13 + 6}2$

$= \frac {12*13 + 6*13 + 2*13 + 6}2=$

$= 6*13 + 3*13 + 1*13 + 3 = 10*13+3 = A3_{13}$.

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By the way. $176_{13} = 13^2 + 7*13 + 6 = 169+91 + 6= 266_{10}$ and $8089_{13}= 8*13^3 + 8*13 + 9 = 17576 + 104 + 9 = 17689_{10}$ and $A3_{13} = 10*13+3 = 133_{10}$.

And we can verify that $266^2 = 4*17689$ by prime factorizing.

$266 = 2*133=2*7*19$ so $266^2 = 2^2 *7^2 *19^2$.

If we divide $17689$ by $7$ we get $2527$ and if we divide that by $7$ again we get $361$. If we divide $361$ by $19$ we get... well, $361 = 380-19$ so $361\div 19 = 20-1 = 19$. So $17689= 7^2*19^2$ and $4*17689 = 4*7^2*19^2 = 266^2$

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Alternative way of solving $x = \frac {176_{13} \pm \sqrt{176_{13}^2 - 4\times8089_{13} }}2$

Is to note $2|176$ so $4|176^2$. $176 = 13^2 + 7*13 + 6 = 12*13 + 8*13 +6 = 20*13 + 6$ so $\frac{176}2 =10*13 + 3 = A3$

So $176_{13}^2 - 4\times8089_{13} = 4*A3^2 - 4*8089$ and so

$x = \frac {176_{13} \pm \sqrt{176_{13}^2 - 4\times8089_{13} }}2=$

$A3 \pm \sqrt {A3^2 - 8089}$.

Now we can demonstrate that $(10*13 + 3)^2 = 8089_{13}$ which.... is as above.... just a lot shorter.

Answer 2

Of course you can change it into:

$$x^2 - 266x + 17689 = 0$$ Which can be solved quite easily, getting $x = 133$

But, any operations you can do in base 10 you can do in base 13 as well.

Using the quadratic formula, you get (all numbers from this point onwards are in base 13):

$$\frac{176 \pm \sqrt{176^2-4*8089}}{2}$$ $$= \frac{176 \pm \sqrt{2628A-2628A}}{2}$$ Interestingly the determinant is zero so it can be easily simplified into: $$ = \frac{176}{2}$$ Which is $A3_{13}$. Checking with the answer in base 10, they have the same result.