The polynomial p(x) = x^2+ax+b has distinct roots 2a and b. Find a+b.
your solution is wrong, since
x^2 + 1/2 x + 1 has complex roots
Instead, we have the product of the roots is
2ab = b
so a = 1/2
and the sum of the roots is
2a+b = -a
b = -3/2
p(x) = x^2 + 1/2 x - 3/2 = 1/2 (x-1)(2x+3)
so the roots are 2a=1 and b = -3/2
a+b = -1
You are correct, thank you for pointing that out! I apologize for my mistake. Your solution is indeed the correct one.
To find the sum a+b, we need to consider the sum of the roots of the polynomial p(x) = x^2 + ax + b.
We are given that the roots are 2a and b.
The sum of the roots can be found by using Vieta's formulas, which state that the sum of the roots of a quadratic equation ax^2 + bx + c = 0 is equal to -b/a.
In this case, the sum of the roots is (2a) + b.
So, we have:
(2a) + b = -a
Simplifying the equation, we get:
3a + b = 0
Since we know that the roots are distinct, this means that a and b cannot be equal to zero simultaneously.
Therefore, from the equation 3a + b = 0, we can conclude that a = 0 and b = 0.
Hence, the sum a + b is 0 + 0 = 0.
Since the roots of the polynomial are 2a and b, we can write:
p(x) = (x - 2a)(x - b)
Expanding the right side:
p(x) = x^2 - bx - 2ax + 2ab
Comparing coefficients with the given polynomial p(x) = x^2 + ax + b, we get:
-b - 2a = a
2ab = b
Adding b to both sides of the second equation, we get:
2ab + b = b(2a + 1)
Since the roots are distinct, we have:
2a ≠ b
Hence, we can divide both sides of the second equation by b to get:
2a = 1
Therefore, a = 1/2 and b = 2a = 1.
Thus, a + b = 1/2 + 1 = 3/2.