4 3 x 2 + 5 x − 2 3 = 0 Δ = 5 2 − 4 ∗ 4 3 ∗ ( − 2 3 ) Δ = 25 + 96 = 121 x = 2 ∗ 4 3 − 5 ± 121 = 8 3 − 5 ± 11 x 1 = 8 3 − 5 − 11 = − 8 3 16 = − 24 16 3 = − 3 2 3 x 2 = 8 3 − 5 + 11 = 8 3 6 = 24 6 3 = 4 3
The zeroes of the polynomial f ( x ) = 4 3 x 2 + 5 x − 2 3 are x 1 = − 3 2 3 and x 2 = 4 3 . The sum and product of the zeroes verify the relationships with the coefficients, confirming that x 1 + x 2 = − a b and x 1 ⋅ x 2 = a c .
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