If a²-4a+1=0 , then the value of a²+a+1/a+1/(a²) is:

If $a^2-4a+1=0$ , then the value of $a^2+a+\dfrac{1}{a}+\dfrac{1}{a^2}$ is:

Show answer & explanation

Correct answer: C

Explanation

The correct answer is option 3 i.e. 18

Explanation :

According to the given question,

⇒ a² - 4a + 1 = 0

Taking a commn in the above equation,

⇒ a(a - 4 +1/a) = 0

⇒ a + 1/a = 4

Also, squaring both sides,

⇒ (a + 1/a)² = 16

⇒ a² + 1/a² + 2 = 16

⇒ a² + 1/a² = 14

Putting these values in the given equation,

⇒ a² + a + 1/a + 1/a² = (a + 1/a) + (a² + 1/a²)

⇒ 4 + 14 = 18.

Hence the required value is 18.

SSC CHSL 2022 PYQ — Question 52