Question

The polynomial p(x) = x^2+ax+b has distinct roots 2a and b. Find a+b.

Answer 1

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

Answer 2

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.

Answer 3

You are correct, thank you for pointing that out! I apologize for my mistake. Your solution is indeed the correct one.