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Mathematics

Solve :

(x2 - x)2 + 5(x2 - x) + 4 = 0

Quadratic Equations

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Answer

Let x2 - x = a

Substituting value in (x2 - x)2 + 5(x2 - x) + 4 = 0 we get,

⇒ a2 + 5a + 4 = 0

⇒ a2 + 4a + a + 4 = 0

⇒ a(a + 4) + 1(a + 4) = 0

⇒ (a + 1)(a + 4) = 0

⇒ a + 1 = 0 or a + 4 = 0      [Zero product rule]

⇒ a = -1 or a = -4

∴ x2 - x = -1 and x2 - x = -4

Solving, x2 - x = -1

⇒ x2 - x = -1

⇒ x2 - x + 1 = 0

Comparing above equation with ax2 + bx + x = 0 we get,

a = 1, b = -1, c = 1

Discriminant = D = b2 - 4ac = (-1)2 - 4(1)(1) = 1 - 4 = -3.

-3 < 0

∴ No real solution.

Solving, x2 - x = -4

⇒ x2 - x + 4 = 0

Comparing above equation with ax2 + bx + x = 0 we get,

a = 1, b = -1, c = 4

Discriminant = D = b2 - 4ac = (-1)2 - 4(1)(4) = 1 - 16 = -15.

-15 < 0

∴ No real solution.

Hence, there is no real solution.

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