Question

(x² + x)^2-(x² + x) - 2 = 0​

Answer 1

Required Answer:-

Given:

• (x² + x)² - (x² + x) - 2 = 0

To find:

• The values of x.

Solution:

We have,

➡ (x² + x)² - (x² + x) - 2 = 0

Let y = x² + x

➡ y² - y - 2 = 0

➡ y² - 2y + y - 2 = 0

➡ y(y - 2) + 1(y - 1) = 0

➡ (y + 1)(y - 2) = 0

Substitute y = x² + x,

➡ (x² + x + 1)(x² + x - 2) = 0

Therefore, either x² + x + 1 = 0 or x² + x - 2 = 0

For the first case,

➡ x² + x + 1 = 0

Here,

★ a = 1

★ b = 1

★ c = 1

Hence,

x = [-b ± √(b² - 4ac)]/2a

= [-1 ±√(1² - 4)]/2

= [-1 ±√-3]/2

= (-1 ±i√3)/2

Hence,

α = (-1 + i√3)/2

β = (-1 - i√3)/2

Again, for the second case,

➡ x² + x - 2 = 0

➡ x² - x + 2x - 2 = 0

➡ x(x - 1) + 2(x - 1) = 0

➡ (x + 2)(x - 1) = 0

Therefore, either x + 2 = 0 or x - 1 = 0

➡ α = -2

➡ β = 1

Hence, the roots of the given quartic equation are (-1 + i√3)/2, (-1 - i√3)/2, -2 and 1.

Answer:

• The roots of the given quartic equation are (-1 + i√3)/2, (-1 - i√3)/2, -2 and 1.