Final answer:
The sum of the reciprocals of the roots of the polynomial 2x²-7x+3 can be calculated using the relationships between roots and coefficients of a quadratic, yielding a value of 7/3.
Step-by-step explanation:
If alpha and beta are the zeroes of the polynomial 2x²-7x+3, we can find the value of 1/alpha + 1/beta using the relationship between the roots and coefficients of a quadratic equation.
To find the value of 1/alpha + 1/beta given that alpha and beta are the zeroes of the polynomial 2x² - 7x + 3, we can use the fact that the sum of the roots of a quadratic equation is equal to the negation of the coefficient of the linear term divided by the coefficient of the quadratic term.
In this case, the sum of the roots is 7/2 and the product of the roots is 3/2. Therefore, 1/alpha + 1/beta = (alpha + beta) / (alpha * beta) = (7/2) / (3/2) = 7/3.
Recall that for a quadratic equation of the form ax² + bx + c = 0, if the roots are alpha and beta, then alpha + beta = -b/a and alpha * beta = c/a. Using these relationships, we can express 1/alpha + 1/beta as:
(1/alpha + 1/beta) = (beta + alpha) / (alpha * beta)
Plugging in the values from our given equation 2x²-7x+3 where a=2, b=-7, c=3, we calculate:
(1/alpha + 1/beta) = (-(-7)/2) / (3/2)
(1/alpha + 1/beta) = (7/2) / (3/2)
(1/alpha + 1/beta) = 7/3
Hence, the value of 1/alpha + 1/beta is 7/3.