This question was previously asked in

UPSSSC Chakbandi Lekhpal Official Paper 3 (Held on : 1 Oct 2019 Shift 1)

Option 4 : 10

__Shortcut Trick__

If, a + 1/a = c

Then, a^{2} + (1/a)^{2} = c^{2} - 2

As, x2 - 3x + 1 = 0

⇒ x(x - 3 + 1/x) = 0

⇒ x + 1/x = 3

Now,

⇒ x^{2} + (1/x)^{2} = 9 - 2

⇒ x2 + (1/x)2 = 7

x + 1/x + x2 + (1/x)2 = 3 + 7

⇒ x + 1/x + x2 + (1/x)2 = 10

**∴ The value of x2 + x + (1/x) + (1/x2) is 10.**

__Additional Information__

** **If, a + 1/a = c

1) a2 + (1/a)2 = c2 - 2

2) a^{2} - (1/a)^{2} = c2 + 2

3) a^{3} + (1/a)^{3} = c^{3} - 3c

4) a3 - (1/a)3 = c^{3} + 3c

**Given:**

x2 - 3x + 1 = 0

**Formula used:**

(a + b)^{2} = a^{2} + b^{2} + 2ab

**Calculation:**

x2 - 3x + 1 = 0

⇒ x(x - 3 + 1/x) = 0

⇒ x - 3 + 1/x = 0

⇒ x + 1/x = 3 ----(1)

After squaring both side of the above equation,

⇒ (x + 1/x)^{2} = 9

⇒ x^{2} + (1/x)^{2} + 2 = 9

⇒ x2 + (1/x)2 = 7 ----(2)

Adding the both equation (1) and equation (2), we get

⇒ x + 1/x + x2 + (1/x)2 = 3 + 7

⇒ x + 1/x + x2 + (1/x)2 = 10

**∴ The value of x2 + x + (1/x) + (1/x2) is 10.**