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How to calculate this using a quadratic equation?

1.56= (x+0)(x+0) / (2-x)(1-x)

User James A
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1 Answer

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Answer:

x = ((18 sqrt(755833) - 17050)^(1/3) - (284 (-1)^(2/3))/(8525 - 9 sqrt(755833))^(1/3))/(15 2^(2/3)) + 1/3 or x = 1/3 + 142/15 ((-2)/(8525 - 9 sqrt(755833)))^(1/3) - 1/15 ((-1)/2)^(1/3) (9 sqrt(755833) - 8525)^(1/3) or x = 1/3 - (2^(1/3) (8525 - 9 sqrt(755833))^(2/3) + 284)/(15 2^(2/3) (8525 - 9 sqrt(755833))^(1/3))

Explanation:

Solve for x over the real numbers:

1.56 = ((x + 0) (x + 0) (1 - x))/(2 - x)

1.56 = 39/25 and ((x + 0) (x + 0) (1 - x))/(2 - x) = (x^2 (1 - x))/(2 - x):

39/25 = (x^2 (1 - x))/(2 - x)

39/25 = ((1 - x) x^2)/(2 - x) is equivalent to ((1 - x) x^2)/(2 - x) = 39/25:

(x^2 (1 - x))/(2 - x) = 39/25

Cross multiply:

25 x^2 (1 - x) = 39 (2 - x)

Expand out terms of the left hand side:

25 x^2 - 25 x^3 = 39 (2 - x)

Expand out terms of the right hand side:

25 x^2 - 25 x^3 = 78 - 39 x

Subtract 78 - 39 x from both sides:

-25 x^3 + 25 x^2 + 39 x - 78 = 0

Multiply both sides by -1:

25 x^3 - 25 x^2 - 39 x + 78 = 0

Eliminate the quadratic term by substituting y = x - 1/3:

78 - 39 (y + 1/3) - 25 (y + 1/3)^2 + 25 (y + 1/3)^3 = 0

Expand out terms of the left hand side:

25 y^3 - (142 y)/3 + 1705/27 = 0

Divide both sides by 25:

y^3 - (142 y)/75 + 341/135 = 0

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

341/135 - 142/75 (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 - 142/75) + (341 z^3)/135 + z^2 (3 λ^2 - (142 λ)/75) + λ^3 = 0

Substitute λ = 142/225 and then u = z^3, yielding a quadratic equation in the variable u:

u^2 + (341 u)/135 + 2863288/11390625 = 0

Find the positive solution to the quadratic equation:

u = (9 sqrt(755833) - 8525)/6750

Substitute back for u = z^3:

z^3 = (9 sqrt(755833) - 8525)/6750

Taking cube roots gives (9 sqrt(755833) - 8525)^(1/3)/(15 2^(1/3)) times the third roots of unity:

z = (9 sqrt(755833) - 8525)^(1/3)/(15 2^(1/3)) or z = -1/15 (-1/2)^(1/3) (9 sqrt(755833) - 8525)^(1/3) or z = ((-1)^(2/3) (9 sqrt(755833) - 8525)^(1/3))/(15 2^(1/3))

Substitute each value of z into y = z + 142/(225 z):

y = 1/15 ((9 sqrt(755833) - 8525)/2)^(1/3) - 142/15 (-1)^(2/3) (2/(8525 - 9 sqrt(755833)))^(1/3) or y = 142/15 ((-2)/(8525 - 9 sqrt(755833)))^(1/3) - 1/15 ((-1)/2)^(1/3) (9 sqrt(755833) - 8525)^(1/3) or y = 1/15 (-1)^(2/3) ((9 sqrt(755833) - 8525)/2)^(1/3) - 142/15 (2/(8525 - 9 sqrt(755833)))^(1/3)

Bring each solution to a common denominator and simplify:

y = ((18 sqrt(755833) - 17050)^(1/3) - (284 (-1)^(2/3))/(8525 - 9 sqrt(755833))^(1/3))/(15 2^(2/3)) or y = 142/15 ((-2)/(8525 - 9 sqrt(755833)))^(1/3) - 1/15 ((-1)/2)^(1/3) (9 sqrt(755833) - 8525)^(1/3) or y = -(2^(1/3) (8525 - 9 sqrt(755833))^(2/3) + 284)/(15 2^(2/3) (8525 - 9 sqrt(755833))^(1/3))

Substitute back for x = y + 1/3:

Answer: x = ((18 sqrt(755833) - 17050)^(1/3) - (284 (-1)^(2/3))/(8525 - 9 sqrt(755833))^(1/3))/(15 2^(2/3)) + 1/3 or x = 1/3 + 142/15 ((-2)/(8525 - 9 sqrt(755833)))^(1/3) - 1/15 ((-1)/2)^(1/3) (9 sqrt(755833) - 8525)^(1/3) or x = 1/3 - (2^(1/3) (8525 - 9 sqrt(755833))^(2/3) + 284)/(15 2^(2/3) (8525 - 9 sqrt(755833))^(1/3))

User VRage
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