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precal

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Solve for y:
(y-(10/y))^2+6(y-(10/y))-27=0

  • precal - ,

    (y-(10/y))^2+6(y-(10/y))-27=0
    his is just like the other one: Let u=y-1/y and you have

    u^2 + 6u -27 = 0
    (u+9)(u-3) = 0
    (y - 10/y + 9)(y - 10/y - 3) = 0
    (y^2 + 9y - 10)(y^2 - 3y - 10) = 0
    (y+10)(y-1)(y-5)(y+3) = 0

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