Solve Quadratic Equation Without Computer

[Albert1]

$\dfrac {1234567891011121314151617}
{7161514131211101987654321}=0.abc----$

please find :$a^2+b^2+c^2=? $

(use of computer is not allowed!)

 

[Opalg]

Re: find :a^2+b^2+c^2=?

$\dfrac {1234567891011121314151617}
{7161514131211101987654321}=0.abc----$

please find :$a^2+b^2+c^2=? $

(use of computer is not allowed!)

[sp]Reminds me of http://mathhelpboards.com/pre-algebra-algebra-2/when-0-123-495051-0-515049-321-a-6136.html. First step is to check that numerator and denominator have the same number of digits (25). Then my little pocket calculator (assuming that I'm allowed to use it) gives the approximations as $$\frac{1234}{7161}\approx 0.1723,$$ $$\frac{12345678}{71615141} \approx 0.17238.$$ It looks as though the first three digits are $1,\ 7,\ 2$, with $1^2+7^2+2^2 = 54.$ I have not tried to prove this carefully as in that previous thread.[/sp]

 

[Albert1]

Re: find :a^2+b^2+c^2=?

$\dfrac {1234567891011121314151617}
{7161514131211101987654321}=0.abc----$

please find :$a^2+b^2+c^2=? $

(use of computer is not allowed!)

Let: $\dfrac{1234}{7162}<A=\dfrac {1234567891011121314151617}
{7161514131211101987654321}=0.abc----<\dfrac{1235}{7160}$

$0.1722<A<0.1725$

$\therefore a=1,b=7,c=2$

and $ a^2+b^2+c^2=54$