If alpha& beta are the zeroes of the polynomial 2x2-7x+3. Find the sum of the reciprocal of its zeroes
this answer is correct.
Why did the polynomial go to therapy? Because it had a constant need for validation from its zeroes! Now, the sum of the reciprocals of the zeroes (alpha and beta) can be obtained by taking the sum of the zeroes and dividing it by the product of the zeroes. In this case:
Sum of reciprocals = (1/alpha) + (1/beta)
= (beta + alpha) / (alpha * beta)
Since we know that alpha and beta are the zeroes of the polynomial, we can substitute them in:
Sum of reciprocals = (beta + alpha) / (alpha * beta)
= (alpha + beta) / (alpha * beta)
Therefore, the sum of the reciprocals of the zeroes is (alpha + beta) / (alpha * beta). I hope this answer adds a little humor to your math journey!
To find the sum of the reciprocals of the zeroes of a polynomial, you need to find the zeroes first. In this case, let's find the zeroes of the polynomial 2x^2 - 7x + 3.
The polynomial can be factored as (2x - 3)(x - 1) = 0.
Setting each factor equal to zero, we get:
2x - 3 = 0 --> x = 3/2
x - 1 = 0 --> x = 1
So, the zeroes of the polynomial are x = 3/2 and x = 1.
Now, to find the sum of the reciprocals of the zeroes, we take the reciprocal of each zero and then add them together:
1/(3/2) + 1/1 = 2/3 + 1 = 2/3 + 3/3 = 5/3
Therefore, the sum of the reciprocals of the zeroes is 5/3.
You are probably dealing with the properties of the sum and product of roots of quadratic equations.
I will use a and b instead of alpha and beta
for 2x^2 - 7x + 3 = 0
the sum of the roots
= a+b
= 7/2
product = ab
= 3/2
so the reciprocals of the roots are 1/a and 1/b
new sum = 1/a + 1/b = (a+b)/(ab)
= (7/2) / (3/2) = 7/3
new product = (1/a)(1/b) = 1/(ab)
= 2/3
new equation:
x^2 - 7/3x + 2/3 = 0
or
3x^2 - 7x + 2 = 0
looking back, looks like I could have stopped after finding the new sum