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    Roots of a quadratic equation

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    Explain why 2 and 3/2 cannot be the roots of 2y2 - 7y + 3 = 0

    © BrainMass Inc. brainmass.com December 24, 2021, 5:05 pm ad1c9bdddf
    https://brainmass.com/math/basic-algebra/roots-quadratic-equation-27417

    SOLUTION This solution is FREE courtesy of BrainMass!

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    If 2 and 3/2 are the roots of a second degree polynomial P(y) we could have written the polynomial as:

    P(y)=(y-2)(y-3/2)

    which gives:

    P(y) = y^2-2y-3y/2+3

    P = (y^2) - (7y/2) + 3

    And this is NOT the original polynomial:

    Q(y) = 2y^2 - 7y + 3

    Thus, 2 and 3/2 are not the roots of Q(x).

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 5:05 pm ad1c9bdddf>
    https://brainmass.com/math/basic-algebra/roots-quadratic-equation-27417

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