# Roots of a quadratic equation

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

Â© BrainMass Inc. brainmass.com December 24, 2021, 5:05 pm ad1c9bdddfhttps://brainmass.com/math/basic-algebra/roots-quadratic-equation-27417

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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).

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