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Find the zeroes of the quadratic polynomial 6x² - 13x +6 and verify the relation between the zeroes and its coefficients.

1 Answer

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We have,

6x {}^(2) - 13x + 6 = 6x {}^(2) - 4x - 9x + 6 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 2x(3x - 2) - 3(3x - 2) = (3x - 2)(2x - 2)

So, to find zeroes of polynomial : 6x² - 13x +6 will be 0,hence ( 3x-2)= 0 and (2x-3) = 0.

So,
x = (2)/(3) \: \: and \: x = (3)/(2)

Therefore, the zeroes of 6x² - 13x +6 are
(2)/(3 ) \: and \: (3)/(2)

Sum of the zeroes

= (2)/(3) + (3)/(2) = (13)/(6) = ( - ( - 13))/(2) = \frac{ - coefficient \: of \: x}{coefficient \: of \: x {}^(2) }

Product of zeroes

= (2)/(3) * (3)/(2) = (6)/(6) = \frac{constant \: term}{coefficient \: of \: x {}^(2) }

Hence verified.
User Eugene Tusmenko
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