Question

Please solve and answer this question fast. Find the roots of the quadratic equation :
2(x+1/x)^2 - 3(x-1/x) - 8 = 0.

Answer 1

Answer:

1,-1,-1/2,2

Step-by-step explanation:

Given Equation is 2(x + 1/x)² - 3(x - 1/x) - 8 = 0

⇒ 2(x² + 1/x)² - 3(x² - 1/x) - 8 = 0

LCM = x²

⇒ 2(x² + 1)² - 3x(x² - 1) - 8x² = 0

⇒ 2(x⁴ + 1 + 2x²) - 3x³ + 3x - 8x² = 0

⇒ 2x⁴ + 2 + 4x² - 3x³ + 3x - 8x² = 0

⇒ 2x⁴ - 3x³ - 4x² + 3x + 2 = 0

⇒ 2x⁴ - x³ - 2x³ - 5x² + x² - 2x + 5x + 2 = 0

⇒ 2x⁴ - x³ - 5x² - 2x - 2x³ + x² + 5x + 2 = 0

⇒ x(2x³ - x² - 5x - 2) - (2x³ - x² - 5x - 2) = 0

⇒ (x - 1)(2x³ - x² - 5x - 2) = 0

⇒ (x - 1)(2x³ + 2x² - 3x² - 3x - 2x - 2) = 0

⇒ (x - 1)(2x²(x + 1) - 3x(x + 1) - 2(x + 1)) = 0

⇒ (x - 1)(x + 1)(2x² - 3x - 2) = 0

⇒ (x - 1)(x + 1)(2x² - 4x + x - 2) = 0

⇒ (x - 1)(x + 1)(2x(x - 2) + (x - 1)) = 0

⇒ (x - 1)(x + 1)(2x + 1)(x - 2) = 0

⇒ x = 1,-1,-1/2,2.

Therefore, the roots of the quadratic equation is 1,-1,-1/2,2.

Hope it helps!

Answer 2

Step-by-step explanation:

2(x + 1/x)² - 3(x - 1/x) - 8 = 0

2(x² + 1/x)² - 3(x² - 1/x) - 8 = 0

2(x² + 1)² - 3x(x² - 1) - 8x² = 0

2(x⁴ + 1 + 2x²) - 3x³ + 3x - 8x² = 0

2x⁴ + 2 + 4x² - 3x³ + 3x - 8x² = 0

2x⁴ - 3x³ - 4x² + 3x + 2 = 0

2x⁴ - x³ - 2x³ - 5x² + x² - 2x + 5x + 2 = 0

2x⁴ - x³ - 5x² - 2x - 2x³ + x² + 5x + 2 = 0

x(2x³ - x² - 5x - 2) - (2x³ - x² - 5x - 2) = 0

(x - 1)(2x³ - x² - 5x - 2) = 0

(x - 1)(2x³ + 2x² - 3x² - 3x - 2x - 2) = 0

(x - 1)(2x²(x + 1) - 3x(x + 1) - 2(x + 1)) = 0

(x - 1)(x + 1)(2x² - 3x - 2) = 0

(x - 1)(x + 1)(2x² - 4x + x - 2) = 0

(x - 1)(x + 1)(2x(x - 2) + (x - 1)) = 0

(x - 1)(x + 1)(2x + 1)(x - 2) = 0

x = 1,-1,-1/2,2.