If $$x^2 + x = 19$$, then what is the value of $$(x + 5)^2+[\frac{1}{(x + 5)^2}]$$ ?

To find : $$(x + 5)^2+[\frac{1}{(x + 5)^2}]$$

Let $$(x+5)=a$$

=> $$x=(a-5)$$ ------------(i)

Thus, we need to find : $$a^2+\frac{1}{a^2}$$ -----------(ii)

Given : $$x^2 + x = 19$$

Substituting value from equation (i),

=> $$(a-5)^2+(a-5)=19$$

=> $$a^2-10a+25+a-5-19=0$$

=> $$a^2-9a+1=0$$

=> $$a^2+1=9a$$

=> $$\frac{a^2+1}{a}=9$$

=> $$a+\frac{1}{a}=9$$

Squaring both sides, we get :

=> $$(a+\frac{1}{a})^2=(9)^2$$

=> $$a^2+\frac{1}{a^2}+2(a)(\frac{1}{a})=81$$

=> $$a^2+\frac{1}{a^2}=81-2=79$$

=> Ans - (B)