0

Before all, good morning; I have just seen an exercise of number systems and equations... I hope you could help me with this:

An equation is given: $$6x^2+60x+150=0.$$ The exercise says that this equation has two real solutions: $$x_1=5, x_2=10.$$

The question is: 

 In which numerical system are the said solutions right (true)?

I have checked in many books of digital systems (I am a student of electronic engineering and that exercise was given in the said subject), but I have not found anything related...

Also, do you know where can I find more exercises like this?

Thanks

Ludolila
  • 3,034
Junjiro
  • 187
  • 1
    How can a positive number like $5$ be a solution? All coefficients of the polynomial on the left side are positive. Maybe the exercise should say it isn't an equation over the reals... – coffeemath Sep 10 '15 at 17:18
  • I suggest a couple approaches to the problem, but none leads to a satisfying solution. I would like to see the intended solution if you find it. Are you sure you have the problem correct? – Ross Millikan Sep 11 '15 at 13:43

1 Answers1

0

I'm surprised to see this in an electrical engineering class. The two number systems that I would think about would be other bases than $10$ and modular arithmetic. As coffeemath points out, you can't have positive roots in any positive base, as all the terms will be positive. We could try negative bases. Let $b$ be the base. Then we must have $6\cdot 5^2 + 6\cdot 5b +b^2+5b=0$ or $b^2+35b+150=0$ where the coefficients are in base $10$, which is satisfied at $b=-5,-30$ We must also have $6b^2+6b^2+b^2+5b=0$ or $13b^2+5b=0$, so no negative base will work for both roots.

For modular arithmetic, where now I assume the coefficients and roots are given in base $10$, we must have $600 \equiv 0$ for the root at $5$ and $1350 \equiv 0$ for the root at $10$. The GCD of these is $50$, so any modulus that is a factor of $50$ will work. Maybe you should insist the modulus be greater than $10$ so that there is not a smaller representative. In that case moduli of $25,50$ are acceptable. Unfortunately, there are other roots, like $15,20$ in these cases.

Ross Millikan
  • 374,822