JOHN
Lv 7
JOHN asked in Science & MathematicsMathematics · 3 years ago

If x and y are the real roots of x³ - 3x² + 5x -17 = 0, and y³ - 3y² + 5y + 11 = 0., find x + y.?

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  • atsuo
    Lv 6
    3 years ago
    Favourite answer

    Let x = u + 1 .

    Given equation x^3 - 3x^2 + 5x - 17 = 0 becomes

    (u + 1)^3 - 3(u + 1)^2 + 5(u + 1) - 17 = 0

    u^3 + 2u - 14 = 0 ---(#1)

    This equation has only one real root .

    Let y = v + 1 .

    Another given equation y^3 - 3y^2 + 5y + 11 = 0 becomes

    (v + 1)^3 - 3(v + 1)^2 + 5(v + 1) + 11 = 0

    v^3 + 2v + 14 = 0 ---(#2)

    This equation has only one real root , too .

    Multiply both sides of (#2) by -1 ,

    -v^3 - 2v - 14 = 0

    (-v)^3 + 2(-v) - 14 = 0 ---(#3)

    This is the same equation as (#1) , so if w is a real root of (#1) then -w is a real root of (#2) .

    That is , x = w + 1 and y = -w + 1 are real roots of two given equations .

    Therefore , x + y = (w + 1) + (-w + 1) = 2 .

  • 3 years ago

    x^3-3x^2+5x-17=0 has one real root & 2 non real roots which are

    x=3.134885

    x=-0.06744242+/-2.327724 i

    y^3-3y^2+5y+11=0 has one real root & 2 non real roots too which are

    y=-1.134885

    y=2.067442+/-2.327724 i

    approximately.

    =>

    x+y=3.134885-1.134885=2

  • iceman
    Lv 7
    3 years ago

    x³ - 3x² + 5x -17 + y³ - 3y² + 5y + 11 = 0

    (x + y - 2) (x² - xy - x + y² - y + 3) = 0

    x + y = 2

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