Final answer:
To find the zeroes of the quadratic polynomial 6x²+x-2, we use the quadratic formula and determine the zeroes to be 1/2 and -3/2. We then verify the relationship between the zeroes and the coefficients of the polynomial, supporting it with Viète's formulas for the sum and product of roots.
Step-by-step explanation:
To find the zeroes of the quadratic polynomial 6x²+x-2, we can use the quadratic formula, which is derived from the equation ax²+bx+c = 0. The zeroes, or roots, of the quadratic are given by:
x = √{-b ± √{b²-4ac}}/{2a}
In this case, a = 6, b = 1, and c = -2. Plugging these into the quadratic formula, we get:
x = √{-1 ± √{(1)²-4(6)(-2)}}/{2(6)}
x = √{-1 ± √{1+48}}/{12}
x = √{-1 ± √{49}}/{12}
x = √{-1 ± 7}/{12}
So the two solutions are:
- x = (√{-1 + 7})/{12} = 1/2
- x = (√{-1 - 7})/{12} = -3/2
To verify the relationship between the zeroes and coefficients, we check the sum and product of the roots. According to Viète's formulas, the sum of the roots α and β is √{-b/a} and their product is c/a.
For the given quadratic:
Sum = α + β = 1/2 + (-3/2) = -1/6 = √{-b/a}
Product = α * β = (1/2) * (-3/2) = -3/4 = c/a
Here, √{-1}/{6} and -2/{6} (or -3/4) correspond with the coefficients of the quadratic polynomial, confirming the relationship between zeroes and coefficients.