Question

Solve:

(i) (x² – x)² + 5 (x² – x) + 4 = 0

(ii) (x² – 3x)² − 16(x² – 3x) – 36 = 0

Answer 1

(i) (x² – x)² + 5 (x² – x) + 4 = 0

Let x² – x = y

Then y² + 5y + 4 = 0

⟹ y² + 4y + y + 4 = 0

⟹ y (y + 4) + 1 ( y + 4) = 0

⟹ (y + 4) ( y + 1) = 0

If y + 4 = 0 Or y + 1 = 0

⟹ x² – x + 4 = 0 Or x² – x + 1 = 0

∴ Given equation has no real solution

(ii) (x² – 3x)² − 16(x² – 3x) – 36 = 0

Let x² – 3x = y

Then y² − 16y − 36 = 0

⟹y² − 18y + 2y − 36 = 0

⟹ y (y − 18) + 2 ( y − 18) = 0

⟹ (y − 18) ( y + 2) = 0

If y − 18 = 0 Or y + 2 = 0

⟹ x² – 3x − 18 = 0 Or x² – 3x + 2 = 0

⟹ x² – 6x + 3x − 18 = 0 Or x² – 2x − x + 2 = 0

⟹ x(x – 6) + 3(x – 6) = 0 Or x (x – 2) – 1 (x – 2) = 0

⟹ (x – 6) (x + 3) = 0 Or (x – 2) (x – 1) = 0

If x – 6 = 0 Or x + 3 = 0 Or x – 2 = 0 Or x − 1 = 0

Then x = 6 Or x = − 3 Or x = 2 Or x = 1