(a) Solve `x^(2) - 15x+ 26 = 0`
(b) Solve `x+1/x = 5/2`.

Explanation

(a) First, let us resolve `x^(2) - 15x + 26` into factors.
`rArr x^(2) - 15x + 26`
`= x^(2) - 13x - 2x + 26`
`= x(x-13) - 2(x-13)`
`= (x-13)(x-2)`.
The given equation, `x^(2) - 15 x +26 = 0` is reduced to `(x-13)(x-2) = 0`
`rArr x - 13= 0` (or) `x - 2 = 0`
`rArr x = 13` (or) x = 2.
`:. x = 2. 13` are the roots of the given equation.
(b) `(x^(2) + 1)/(x) = 5/2`
`rArr 2x^(2) + 2 = 5x`.
`rArr 2x^(2) - 5x + 2 = 0`
`rArr 2x^(2) -4x - x + 2 = 0`
`rArr 2x(x-2) - 1 (x-2)= 0`
`rArr (2x-1)(x-2) = 0`
`rArr 2x - 1 = 0` or `x - 2 = 0`
`rArr x = 1/2` or `x = 2`.
`:. x = 1/2, 2` are the roots of the given equation.