# Finding a Digit of One Root of $x\sqrt{8}+\frac{1}{x\sqrt{8}}-\sqrt{8}$ Knowing The Digit of The Other Root

Consider $$f(x)=x\sqrt{8}+\frac{1}{x\sqrt{8}}-\sqrt{8}$$. The function has 2 real roots, say, $$x_1$$ and $$x_2$$. If the $$1994$$th digit in the decimal expansion of $$x_1$$ is $$6$$, what is the $$1994$$th digit in the decimal expansion of $$x_2$$?

This is question 1 of the 34th Swedish Olympiad (1994). Either by explicitly finding the roots, or by Vieta's relations, we get that $$x_1+x_2=1$$. Now, if we write $$1$$ as $$0.999\cdots$$, we see that the corresponding digits in the decimal place add up to $$9$$. That is, the $$m$$th digit of $$x_1$$ plus the $$m$$th digit of $$x_2=9$$, for $$m\geq2$$. This means that the $$1994$$th digit of $$x_2$$ is $$3$$.

Is this correct? For some reason, I can't find the answer/solution anywhere.

Also, if we have $$a,b\in\mathbb{R}$$, and $$(a+b)\in\mathbb{Z}$$, then what can be said about the decimal expansion of $$a$$ and $$b$$? For example, $$\pi=3.1415\cdots$$, and $$1-\pi=-2.1415\cdots$$. We can see that the digits in the decimal expansion of $$\pi$$ and $$1-\pi$$ are equal (except for the first digit, of course). Can this be generalised?

Edit: I thought it would be helpful to include a screenshot that demonstrates that corresponding digits in the decimal expansions of $$x_1$$ and $$x_2$$ add up to $$9$$:

• Your answer looks correct to me. As to generalization, I think the answer would be that either the digits of $a,b$ are the same, or the digits of $b, d_b$, are $9-d_a$. Commented Dec 26, 2022 at 5:25
• An exception must be made for terminating decimals, e.g. $0.63 + 0.37 = 1$. Commented Dec 26, 2022 at 5:33
• but since the roots are non-terminating here, this exception can be overlooked here.
– D S
Commented Dec 26, 2022 at 12:46
• Also, note that this is negative for negative $x$ (all terms are negative). Zero clearly isn't a root. Thus, both roots are positive. Combined with the observation that the roots sum to $1$, this implies they're in the range $(0,1)$. Commented Dec 26, 2022 at 13:48
• @RobertIsrael Technically, even if the 1994th digit onwards is $0.\dots600\dots$, the 1994th digit of the other root will be ambiguous, since it could be written either as $0.\dots400\dots$ or $0.\dots399\dots$. So answering with $3$ would be a valid solution regardless. Commented Dec 26, 2022 at 13:52

$$9-$$nth digit of $$x_{1}=$$ nth digit of $$x_{2}$$. "An exception must be made for terminating decimals, e.g. 0.63+0.37=1 . – Robert Israel Dec 26, 2022 at 5:33" As this is not a terminating decimal, but an irrational number, with $$x_{1}+x_{2}=0.9999999...$$ , Robert Israel's statement will not work here. So the other 1994th digit of the other root should be $$9-3=6$$. For your example of $$\pi$$ and $$1-\pi$$, as the numbers of both roots are positive, that case does not work here. (Try $$4-\pi$$)