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

Question

Mathematics

Howard Cannon

3 August, 10:40

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

+4

Answers (1)

Know the Answer?

Hannah Contreras · Answer 1

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: 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))