Age old problem

[dave7]

Today, a is twice the age of b. A few yrs ago, a's age was 12 times the square of b's present age.What are their ages ? I set up a grid :
a b
2x x
7x ^1/2 y THE X AND Y REPRESENT AGES FOR THE TIME PERIODS. N IS THE TIME DIFFERENCE X - N = Y
2X - N = 7X ^ 1/2. THANKS.

 

[soroban]

Hello, dave7!

I don't understand your equations.
. . Besides the problem has no solution.

Please check for typos.

Today A is twice the age of B.
A few years ago, A's age was 12 times the square of B's present age.
What are their ages?

Let \(\displaystyle x\) = B's present age.

\(\displaystyle \begin{array}{c||c|c|} & \text{Now} & n\text{ yrs ago} \\ \hline \hline
A & 2x & 2x-n \\ \hline B & x & x-n \\ \hline \end{array}\)

The equation is: .\(\displaystyle 2x-n \:=\:12x^2\)

And we have: .\(\displaystyle 12x^2 - 2x + n \:=\:0\)
. . which has no real solution for positive \(\displaystyle n.\)

 

[HallsofIvy]

Yes, thought of as a quadratic Diophantine equation, with integer solutions, that problem has a solution.