Mathematics, 19.05.2020 14:13 hdbug2025

A system of equations consists of y=x^3+5x+1 and y=x

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Answer from: jasmine8142002

x = (2^(1/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(2/3) - 10 3^(1/3))/(6^(2/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3)) or x = -((-1)^(1/3) (2^(1/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(2/3) + 10 (-3)^(1/3)))/(6^(2/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3)) or x = ((-1)^(1/3) ((-2)^(1/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(2/3) + 10 3^(1/3)))/(6^(2/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3))

Step-by-step explanation:

Solve for x:

y = x^3 + 5 x + 1

y = x^3 + 5 x + 1 is equivalent to x^3 + 5 x + 1 = y:

x^3 + 5 x + 1 = y

Subtract y from both sides:

1 + 5 x + x^3 - y = 0

Change coordinates by substituting x = z + λ/z, where λ is a constant value that will be determined later:

1 - y + 5 (z + λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:

z^6 + z^4 (3 λ + 5) + z^3 (1 - y) + z^2 (3 λ^2 + 5 λ) + λ^3 = 0

Substitute λ = -5/3 and then u = z^3, yielding a quadratic equation in the variable u:

-125/27 + u^2 - u (y - 1) = 0

Find the positive solution to the quadratic equation:

u = 1/18 (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))

Substitute back for u = z^3:

z^3 = 1/18 (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))

Taking cube roots gives (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3)/(2^(1/3) 3^(2/3)) times the third roots of unity:

z = (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3)/(2^(1/3) 3^(2/3)) or z = -((-1/2)^(1/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3))/3^(2/3) or z = ((-1)^(2/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3))/(2^(1/3) 3^(2/3))

Substitute each value of z into x = z - 5/(3 z):

x = (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3)/(2^(1/3) 3^(2/3)) - (5 (2/3)^(1/3))/(-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3) or x = -(5 (-1)^(2/3) (2/3)^(1/3))/(-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3) - (((-1)/2)^(1/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3))/3^(2/3) or x = (5 ((-2)/3)^(1/3))/(-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3) + ((-1)^(2/3) (-9 + 9 y + sqrt(3) sqrt(27 y^2 - 54 y + 527))^(1/3))/(2^(1/3) 3^(2/3))

Bring each solution to a common denominator and simplify:

Answer from: Quest

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.

Answer from: Quest

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step-by-step explanation:

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