If \( {x }^{ 2} \) + x = 41
find \( {(x + 7) }^{ 2} \) + \(\frac{1}{(x+7)^2}\)
Answer & explanation
Correct answer: option 3
Let x + 7 = t
x = t - 7
Put in \( {x }^{ 2} \) + x = 41
⇒ (t - 7)2 + t - 7 = 23
⇒ t2 + 49 - 14t + t - 7 = 41
⇒ t2 - 13t + 1 = 0
⇒ t + \(\frac{1}{t}\) = 13
⇒ t2 +\(\frac{1}{ {t }^{2 }}\) = 132 - 2 = 167