Question 750058
For which value satisfies the equation {{{10a^2-b^2=1}}} when a is greater 40 and b is greater than 100?
:
Rearrange to
10a^2 = b^2 + 1
a^2 = {{{(b^2+1)/10}}}
:
a = {{{sqrt((b^2+1)/10)}}} 
let a > 40 
{{{sqrt((b^2+1)/10)}}} > 40
Square both sides
{{{(b^2+1)/10)}}} > 1600
b^2 + 1 > 1600 * 10
b^2 > 16000-1
b > {{{sqrt(15999)}}}
b = 127 
Find a
10a^2 = b^2 + 1
10a^2 = 127^2 + 1
10a^2 = 16129 + 1
a^2 = 16130/10
a^2 = 1613
a = {{{sqrt(1613)}}}
a = 40.162