Quadratic problem: 19 Jan'16

Question

The number of solution of !!10^(2/x) + 25^(1/x) = 65/8(50^(1/x))!!

!!a)" "1!!

!!b)" "2!!

!!c)" "3!!

!!d)" "oo!!

Solution

!!10^(2/x) + 25^(1/x) = 65/8(50^(1/x))!!

!!=> (2xx5)^(2/x) + (5xx5)^(1/x) = 65/8(2xx5xx5)^(1/x)!!

!!=> (2^(2/x))xx(5^(2/x)) + 5^(2/x) = 65/8(2^(1/x))xx(5^(2/x))!!

!!=> (5^(2/x)) [(2^(2/x)) + 1] = 65/8(2^(1/x))xx(5^(2/x))!!

So now, !!(5^(2/x)) = 0!!, this is not possible

and !!(2^(2/x)) + 1 = 65/8(2^(1/x))!!

Putting, !!(2^(1/x)) = z!!

Then, !!z^2 + 1 = 65/8z!!

!!=> 8z^2 + 8 = 65z!!

!!=> 8z^2 - 65z + 8 = 0!!

!!=> 8z^2 - 64z - z + 8 = 0!!

!!=> 8z(z-8) - 1(z-8) = 0!!

!!=> (8z-1)(z-8) = 0!!

So, !!z = 1/8 and 8!!

!!=> 2^(1/x) = 1/8 and 8!!

This will give, !!x = -1/3 and 1/3!!