How to solve a quadratic equation with Binary coefficients?

Question

Consider the quadratic equation: $$5x^2 - 50x + 125 = 0$$ It has the roots $x_1 = 5$ and $x_2 = 5$. But now, convert these coefficients into binary: $$101x^2 - 110010x + 1111101 = 0$$ How can I solve this quadratic equation with binary coefficients?!

This is part of the bigger question: What would have been our number system if humans had more than 10 fingers? Try to solve this puzzle.

but that's just the same equation. The numbers don't change just because you write them differently. — lulu, Sep 12 '20 at 15:24
The same way as you do with decimal coefficients; you end with $x_1=101$ and $x_{10}=101$. — Angina Seng, Sep 12 '20 at 15:24
Solving an equation is independent of your choice of way to represent its coefficients. — José Carlos Santos, Sep 12 '20 at 15:24
Oh okay... the original question is here: https://math.stackexchange.com/questions/460729/what-would-have-been-our-number-system-if-humans-had-more-than-10-fingers-try-t/460737 — Anuj Manoj Shah, Sep 12 '20 at 15:28
The original question was to show was to show that the equation was in base thirteen. — Angina Seng, Sep 12 '20 at 15:38
Ohh okay, the coefficients were themselves in base-13. Thanks guys! — Anuj Manoj Shah, Sep 12 '20 at 15:40

score 1 · Answer 1 · answered Sep 12 '20 at 15:39

Just because the coefficients of the quadratic equation are in a different base doesn't mean that its roots are different! In decimal, the roots are: $$x_1 = 5$$ $$x_2 = 5$$ and because $5 = (101)_2$ in binary, thus the roots of the equation $(101)_2 x^2−(110010)_2 x+(1111101)_2=0$ are: $$x_1 = (101)_2$$ $$x_2 = (101)_2$$

fleablood · Accepted Answer · 2020-09-12T16:29:53.697

Changing the basis of a number system doesn't change the results of the arithmetic; it only changes how we right the the numbers.

if $a=$ then number we think of as five and $b=$ the number we think of as fifty and $c=$ the number we think of one hundred and twenty five then the solution so $ax^2 - bx + c = 0$ is $\pm a$ no matter how we right the numbers.

The solution to $5_{10}x^2 - 50_{10}x +125_{10}=0$ will be $x=\pm 5_{10}$ and the solution to $5_8x^2 - 62_8x + 201_8=0$ will be $x= \pm 5_8$ and the solution to $101_x^2−110010_2x+1111101_2=0$ will be $x = \pm 101_2$.

What the question in the linked question is asking is different though:

We have $5_bx^2 - 50_bx + 125_b = 0$ has solutions $x=5_b; 8_b$. But we don't know what $b$ is or what any of the numbers actually are.

so what is $b$?

Well, the quadratic formula is the quadratic formula so

So if $A = 5_b$ then $b \ge 6$ and $A = 5$ and if $B= 50_b = 5b$ and $C=125_b = b^2 + 2b + 5b$ then the solutions

$\frac {B-\sqrt{B^2 - 4AC}}{2A} = 5$ and $\frac {B+\sqrt{B^2 - 4AC}}{2A} = 8_b = 8$ and $b \ge 9$.

And subtracting those solutions we get $\frac {B+\sqrt{B^2 - 4AC}}{2A}-\frac {B-\sqrt{B^2 - 4AC}}{2A}= 8-5$ so

$\frac {\sqrt{B^2 - 4AC}}{A} = 3$ and as $A=5$

$\sqrt{B^2 - 20C} = 15$ so $B^2 - 20C = 225$.

And $B = 5b$ and $C=b^2 + 2b + 5$ so $25b^2 - 20b^2 - 40b - 100 = 225$

$5b^2 -40b -325=0$

$b^2 - 8b - 65=0$ so $b = \frac {8 \pm \sqrt {64+4*65}}2=4 \pm \sqrt{16+65}=4\pm 9=-5,13$.

so we can conclude $b = 13$.

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Oh for eff's sake:

if the so solutions are $x=5, x=8$ then the equation is $5(x-5)(x-8) = 5(x^2 -13x + 40)$ and $50_b = 5*13$ and $b = 13$.

sheesh that's the the interviewers were going for. And I'd have failed...

unless they want a person who rolls up their sleeves for every problem...

I actually got this question from a book called Digital Design (John F. Wakerly). Specifically, from the chapter on "Number Systems and Codes". — Anuj Manoj Shah, Sep 13 '20 at 13:12

How to solve a quadratic equation with Binary coefficients?

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