Quadratic Equations

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The longer leg of a right triangle is 3 ft longer than three times the shorter leg. The hypotenuse is 3 ft shorter than four times the shorter leg. Find the lengths of the three sides of the right triangle. (Please provide steps/explanation to solve this problem.)

  • Quadratic Equations -

    shoter leg = x
    longer leg = 3x+3
    hypot= 4x-3

    to solve for hypotenuse you use
    a^2 + b^2 = c^2, so

    (x)^2 + (3x+3)^2 = (4x-3)^2

    (x)(x) + (3x+3)(3x+3) = (4x-3)(4x-3)
    then multiply

    (x^2) + (9x^2+18x+9) = (16x^2-24x+9)

    then combine like terms

    10x^2 + 18x + 9 = 16x^2 - 24x + 9

    then i subtracted 10x^2 from both sides
    and got:

    18x + 9 = 6x^2 - 24x +9

    next, i subtracted 18x from both sides:

    9 = 6x^2 - 42x + 9

    and then subtracted 9 from both sides

    0 = 6x^2 - 42x

    then pull out a 6x on the right side of the equation

    0 = 6x(x-7)

    so
    6x=0
    and
    x-7=0

    so x= 0 and 7

    so the shorter leg is :
    0 units or 7 units

    the longer leg is :
    3 units or 24 units

    and the hypotenuse is :
    -3 units or 25

    and since you cant have a hypotenuse with a negative value
    i think the three lengths of your triangle are

    7, 24, and 25

  • Quadratic Equations -

    Thanks for the help. :)

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