0
$\begingroup$

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.

$\endgroup$
  • 6
    $\begingroup$ but that's just the same equation. The numbers don't change just because you write them differently. $\endgroup$ – lulu Sep 12 at 15:24
  • 1
    $\begingroup$ The same way as you do with decimal coefficients; you end with $x_1=101$ and $x_{10}=101$. $\endgroup$ – Angina Seng Sep 12 at 15:24
  • 4
    $\begingroup$ Solving an equation is independent of your choice of way to represent its coefficients. $\endgroup$ – José Carlos Santos Sep 12 at 15:24
  • $\begingroup$ Oh okay... the original question is here: math.stackexchange.com/questions/460729/… $\endgroup$ – Anuj Manoj Shah Sep 12 at 15:28
  • 1
    $\begingroup$ The original question was to show was to show that the equation was in base thirteen. $\endgroup$ – Angina Seng Sep 12 at 15:38
0
$\begingroup$

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 $b = 13$.

=======

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...

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Interesting that $125_{13} = 200_{10}$.... a little... $\endgroup$ – fleablood Sep 12 at 16:32
  • $\begingroup$ I actually got this question from a book called Digital Design (John F. Wakerly). Specifically, from the chapter on "Number Systems and Codes". $\endgroup$ – Anuj Manoj Shah Sep 13 at 13:12
0
$\begingroup$

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: $$x_1 = (101)_2$$ $$x_2 = (101)_2$$

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.