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13 April, 21:59

How to calculate this using a quadratic equation?

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

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  1. 13 April, 22:22
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    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))

    Step-by-step 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))
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