Given the fraction E(x)=(x^3-3x-2)/(x^3+1+3x^2+3x). Solve the equation E(x)=square root 3-square root 2.

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We have E(x) = (x^3-3x-2)/(x^3+1+3x^2+3x). We have to solve E(x) = sqrt 3 - sqrt 2.

E(x) = sqrt 3 - sqrt 2

=> E(x) = (x^3-3x-2)/(x^3+1+3x^2+3x) = sqrt 3 - sqrt 2

=> (x^3-3x-2)/(x + 1)^3 = sqrt 3 - sqrt 2

=> (x^3 - x - 2x -...

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We have E(x) = (x^3-3x-2)/(x^3+1+3x^2+3x). We have to solve E(x) = sqrt 3 - sqrt 2.

E(x) = sqrt 3 - sqrt 2

=> E(x) = (x^3-3x-2)/(x^3+1+3x^2+3x) = sqrt 3 - sqrt 2

=> (x^3-3x-2)/(x + 1)^3 = sqrt 3 - sqrt 2

=> (x^3 - x - 2x - 2)/(x + 1)^3 = sqrt 3 - sqrt 2

=> (x(x^2 - 1) - 2(x + 1)/(x + 1)^3 = sqrt 3 - sqrt 2

=> (x(x - 1)(x + 1) - 2(x + 1))/(x + 1)^3 = sqrt 3 - sqrt 2

=> (x + 1)( x^2 - x - 2)/(x + 1)^3 = sqrt 3 - sqrt 2

=> (x^2 - x - 2)/(x + 1)^2 = sqrt 3 - sqrt 2

=> (x^2 - 2x + x - 2)/(x + 1)^2 = sqrt 3 - sqrt 2

=> (x(x -2) +1(x - 2)/(x + 1)^2 = sqrt 3 - sqrt 2

=> (x+1)(x -2)/(x + 1)^2 = sqrt 3 - sqrt 2

=> (x -2)/(x + 1) = sqrt 3 - sqrt 2

=> x - 2 = (sqrt 3 - sqrt 2)(x + 1)

=> x - 2 = (x* sqrt 3 - x* sqrt 2 + sqrt 3 - sqrt 2)

=> x - x*sqrt 3 + x*sqtr 2 = 2 + sqrt 3 - sqrt 2

=> x( 1- sqrt 3 + sqrt 2) = 2 + sqrt 3 - sqrt 2

=> x = (2 + sqrt 3 - sqrt 2) / ( 1- sqrt 3 + sqrt 2)

The required value of x = (2 + sqrt 3 - sqrt 2) / ( 1- sqrt 3 + sqrt 2)

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