If x² + x = 19, then find (x + 5)² + \(\frac{1}{(x + 5)^2}\)

79

Let x + 5 = t 

we have to find t² + \(\frac{1}{t^2}\)

x + 5 = t

x = t - 5

Put the value of x in:

x² + x = 19

(t -5)² + (t - 5) = 19

⇒ t² + 25 - 10t + t - 5 = 19

⇒ t² - 9t + 1 = 0

⇒ t + \(\frac{1}{t}\) = 9                   

⇒ t² + \(\frac{1}{t^2}\) = 9² - 2 = 79

{here we use: If x + \(\frac{1}{x}\) = a  , then x² + \(\frac{1}{x^2}\) = a² - 2}