If $2a +\frac{1}{a} = 4$, then the value of $a^2 + \frac{1}{4a^2}$ is :
Answer & explanation
Correct answer: option 1
If $K+\frac{1}{K}=n$
then, $K^2+\frac{1}{K^2}$ = n2 – 2 × k × \(\frac{1}{k}\)
If $2a +\frac{1}{a} = 4$,
Then $4a^2 +\frac{1}{a^2}$ = 42 – 2 × 2a × \(\frac{1}{a}\)
$4a^2 +\frac{1}{a^2}$ = 16 – 2 × 2 = 12
Divide the equation by 4 for getting the desired type of result,
then the value of $a^2 + \frac{1}{4a^2}$ is = \(\frac{12}{4}\) = 3