# How to solve a quadratic equation with Binary coefficients? [closed]

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 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 at 15:24
• Solving an equation is independent of your choice of way to represent its coefficients. – José Carlos Santos Sep 12 at 15:24
• Oh okay... the original question is here: math.stackexchange.com/questions/460729/… – Anuj Manoj Shah Sep 12 at 15:28
• The original question was to show was to show that the equation was in base thirteen. – Angina Seng Sep 12 at 15:38

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...

• Interesting that $125_{13} = 200_{10}$.... a little... – fleablood Sep 12 at 16:32
• 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 at 13:12

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