SSC BOARD PAPERS IMPORTANT TOPICS COVERED FOR BOARD EXAM 2024

### 2(x2 + 1/x2 ) – 9(x+1/x) + 14 = 0

(iii) 2(x2 + 1/x2 ) – 9(x+1/x) + 14 = 0

Sol.  2(x2 + 1/x2 ) – 9(x+1/x) + 14 = 0

∴ 2[(x + 1/x)2 – 2] - 9(x+1/x) + 14 = 0

[∵(x2 + 1/x2 ) = [(x + 1/x)2 – 2]

Put x + 1/x = m,

∴ 2(m2 – 2) – 9m + 14 = 0

∴ 2m2 – 4 – 9m + 14 = 0

∴ 2m2 – 9m + 10 = 0

∴ 2m2 – 4m – 5m + 10 = 0

∴ 2m (m – 2) – 5 (m – 2) = 0

∴  (m – 2) (2m – 5) = 0

∴  m – 2 = 0 or 2m – 5 = 0

∴  m = 2 or 2m = 5

∴ m = 2 or m = 5/2

Resubstituting m = x + 1/x

x + 1/x = 2  ....(i)    x + 1/x = 5/2 ..... (ii)

From (i)

x + 1/x = 2

Multiplying by x, we get

x2 + 1 = 2x

∴ x2 – 2x + 1 = 0

∴ x2 – x – x + 1 = 0

∴ x(x – 1) – 1 (x – 1) = 0

∴ (x – 1) (x – 1) = 0

∴ x – 1 = 0    or   x – 1 = 0

∴ x = 1  or  x  = 1

i.e. x = 1

From  (ii)

x + 1/x = 5/2

Multiplying by 2x, we get

2x2 + 2 = 5x

∴ 2x2 – 5x + 2 = 0

∴ 2x2 - 4x – 1x + 2 = 0

∴ 2x(x – 2) -1 (x – 2) = 0

∴ (2x – 1)(x – 2) = 0

∴ 2x – 1 = 0    or x – 2 = 0

∴ x = ½   or x = 2

∴ x = 1 or ½  or 2