f(x)=2x^2−146x+c has 2 roots that are positive prime numbers. What is c?
To find the value of c in the given quadratic equation f(x) = 2x^2 - 146x + c, we need to use the information that the equation has two roots that are positive prime numbers.
A quadratic equation can be represented in the form ax^2 + bx + c = 0, where a, b, and c are constants.
In this case, we are told that the roots of the equation are positive prime numbers. We know that prime numbers are only divisible by 1 and themselves, and they are greater than 1.
Let's consider the prime roots as p and q.
So, the equation can be rewritten as:
f(x) = 2(x - p)(x - q) = 2x^2 - 2px - 2qx + 2pq
Comparing this with the given equation f(x) = 2x^2 - 146x + c, we can see that:
p + q = -b/a = -(-146)/2 = 73
Also, from the equation f(x) = 2(x - p)(x - q), we know that the constant term is equal to the product of the roots:
c = 2pq
Since the roots p and q are prime numbers, their product 2pq will also be a positive prime number.
To find two prime numbers that add up to 73, we can try different combinations:
1 + 72 = 73 (Not prime)
2 + 71 = 73 (Prime)
3 + 70 = 73 (Not prime)
4 + 69 = 73 (Not prime)
...
After checking various combinations, we find that 2 and 71 are both prime numbers and sum up to 73. Therefore, p = 2 and q = 71.
Now, we can calculate the value of c:
c = 2pq = 2 * 2 * 71 = 284
Hence, c is equal to 284.