Find the sum of natural values of m with which the equation...

\(\frac{t-6}{m-8} = \frac{m}{t}\)
\(t^2-6t=m^2-8m\)

Add some integer values on each to get the form (a-b)^2 on both sides
\(t^2-6t+9-9=m^2-8m+16-16\)
\((t-3)^2-9=(m-4)^2-16\)
\((t-3)^2=(m-4)^2-7\)
Whenever the right hand side is less than 0, we will not have real values as \((t-3)^2\) cannot be negative

So the values of m that satisfy the given condition of no real root=> \((m-4)^2-7<0.........(m-4)^2<7\)
\(|m-4|<\sqrt{7}.......|m-4|<2.6\)
Two cases
1) \(m-4<2.6.......m<6.6\)
2) \(4-m<2.6.......m>1.4\)
Possible integer values of \(m = {2,3,4,5,6}\)
Sum =\(2+3+4+5+6=20\)

OR

You can also substitute value of m in \((m-4)^2<7\)
m cannot be negative or 0 as \((m-4)^2\) will become greater than 7.
Put 1,2,3 and so on to get the values as 2,3,4,5,6

A