What are the roots of: 8*x^4-48*x^3+74*x^2-42*x+8 = 0

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve: 8*x^4 - 48*x^3 + 74*x^2 - 42*x + 8 = 0

8*x^4 - 48*x^3 + 74*x^2 - 42*x + 8 = 0

=> 4*x^4 - 24*x^3 + 37*x^2 - 21*x + 4 = 0

=> 4*x^4 - 16*x^3 - 8x^3 + 32*x^2 + 5*x^2 - 20*x - x  + 4 = 0

=> 4x^3(x - 4) - 8x^2(x - 4) + 5x(x - 4) - 1(x - 4) = 0

=> (x - 4)(4x^3 - 8x^2 + 5x - 1) = 0

=> (x - 4)(4x^2 - 4x^2 - 4x^2 + 4x + x - 1) = 0

=> (x - 4)(4x^2(x - 1) - 4x(x - 1) + 1(x - 1)) = 0

=> (x - 4)(x - 1)(4x^2 - 4x + 1) = 0

=> (x - 4)(x - 1)(4x^2 - 2x - 2x + 1) = 0

=> (x - 4)(x - 1)(2x(2x - 1) - 1(2x - 1)) = 0

=> (x - 4)(x - 1)(2x - 1)( 2x - 1) = 0

=> x = 4, x = 1 , x = 1/2

The solutions of the equation are {1/2 , 1 , 4}

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