solve 2(x + 1/x)^2 - 3(x-1/x)-8=0 Share with your friends Share 7 Mayank Jha answered this Dear student, 2(x + 1x)² - 3(x - 1x) - 8 = 0⇒2*(x² + 1x)² - 3(x² - 1x) - 8 = 0⇒2(x² + 1)²x²-3(x² - 1)x-8 = 0⇒2(x4 + 2x² + 1) - 3x(x² - 1) - 8x² = 0⇒2x4 + 4x² + 2 - 3x3 + 3x - 8x² = 0⇒2x4 - 3x3 - 4x² +3x+ 2 = 0Value of the given polynomial is zero at x = 1. So, (x - 1) is one of the factors.⇒2x3(x - 1)-x²(x - 1)-5x(x - 1)-2(x - 1) = 0⇒(x - 1)(2x3 - x² - 5x - 2) = 0Now, let us consider the polynomial: (2x3 - x² - 5x - 2). (x + 1) will be the factor of this polynomial.(2x3 - x² - 5x - 2)⇒2x²(x + 1)-3x(x + 1)-2(x + 1)⇒(x + 1) (2x² - 3x - 2)⇒(x + 1)(2x² - 4x + x - 2)⇒(x + 1)(2x(x -2) + 1(x - 2))⇒(x + 1)(x - 2)(2x + 1)So, 2x4 - 3x3 - 4x² +3x+ 2 = 0 can be written as:2x4 - 3x3 - 4x² +3x+ 2 = 0⇒(x + 1)(x - 2)(2x + 1)(x - 1) = 0∴x = ±1, 2, -12 Regards -2 View Full Answer
