Question: Find an equation whose roots are the reciprocals of the roots of 2x^2 + x -4 = 0
What i did so far:
sum = -b/a = -1/2 --> reciprocal = -2
product = c/a = -2 --> reciprocal = -1/2
Therefore, x^2 - (sum)x + (product) = x^2 - (-2)x + (-1/2)
= 2x^2 + 4x -1
From other sources, another answered showed: 4x^2 + x -2
Can someone verify whether what i did is correct?
Suppose you have an equation with roots 2 and 3. Then the sum of the roots is 5.
The reciprocals of 2 and 3 are 1/2 and 1/3, and their sum is 5/6, which is not the reciprocal of 5.
So the roots are the reciprocals but the sum of the roots is not a reciprocal.
I would suggest finding the roots of the original equation as it will make your job easier.
Oh right right, my mistake, i think this should be done like:
Normally, sum = a + b, but the reciprocal is:
sum = 1/a + 1/b
sum = (b + a) / ab
b + a is the sum, and sum = -b/a, and therefore is equal to: -1/2
ab is the product, and product = c/a, and therefore is equal to: -2
So sum = (-1/2) / (-2) = 1/4
Normally, product = ab, but the reciprocal is:
product = (1 / a)(1 / b)
= (1 / ab)
= (1 / -2)
Therefore the equation is x^2 + 1/4x - 1/2 = 0 OR 4x^2 - x - 2 = 0