Question : If $x^{2}+x=19,$ then what is the value of $(x+5)^{2}+\left [ \frac{1}{(x+5)^{2}} \right ]\;?$
Option 1: 77
Option 2: 79
Option 3: 81
Option 4: 83
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Correct Answer: 79
Solution :
Given: $x^{2}+x=19$
To find: $(x+5)^{2}+\left [ \frac{1}{(x+5)^{2}} \right ]$......(1)
Let us consider $x+5 = a$ ⇒ $x = a-5$
Putting the value of $x$ in the given equation,
$(a-5)^2 +(a-5) = 19$
⇒ $a^2 +25 - 10a + a - 5-19 = 0$
⇒ $a^2 -9a+1 = 0$
⇒ $a^2+1=9a$
⇒ $a+\frac{1}{a} = 9$
Squaring both sides, we get,
$a^2 + \frac{1}{a^2} + 2= 81$
⇒ $a^2 + \frac{1}{a^2} = 79$
From equation (1),
⇒ $(x+5)^{2}+\left [ \frac{1}{(x+5)^{2}} \right ] = 79$
Hence, the correct answer is 79.
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