The equation x^{2}+ ax + 2023 = 0 has roots x_{1} = b + 1 and x_{2} = 8b^{2} + 1 for some real number b. Find a + b.

So far I have been trying Vieta's and getting 8b^{3} + 8b^{2} + b +1 = 2023, -a = 8b^{2} + b + 2. And then getting 2023 + a = 8b^{3}, but then I don't know what to do from here, is my approach totally wrong? Also, can I have the answer with a solution, so I understand how you got the answer?

Guest May 20, 2023

#1**0 **

We know that the sum of the roots of a quadratic equation is equal to -b/a and the product of the roots is equal to c/a. In this case, the sum of the roots is b + 1 + 8b2 + 1 = 8b2 + b + 2 and the product of the roots is (b + 1)(8b2 + 1) = 8b3 + b + 1. We are given that the sum of the roots is equal to -a/1 and the product of the roots is equal to 2023. Therefore, we have the following equations:

8b2 + b + 2 = -a/1 8b3 + b + 1 = 2023

Solving for a and b, we get a = -8b2 and b = 3. Therefore, a + b = -8b2 + b = -7b2 = -7(3) = -21.

Guest May 21, 2023