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Find the partial acon
x - 3x +1
(x-1)²(x-2)
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student-name Angela Achy Abraham asked in Math
solve 2(x + 1/x)^2 - 3(x-1/x)-8=0
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student-name Mayank Jha answered this
525 helpful votes in Math, Class XII-Science
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
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