Let the equations x2 + x + 2 = 0 and ax2 + bx + 1 = 0 (a, b ∈ R) have atleast one common root, then (a + b) equals

Given that the two equations have at least one common root, we can use the concept of coefficients to solve this problem.

The equations given are:

1. x² + x + 2 = 0
2. ax² + bx + 1 = 0

Let's assume that the common root is 'p'.

Then, by Vieta's formulas, the sum of the roots for the first equation is -b/a = -1/1 = -1. This means that the other root of the first equation is -1 - p.

Similarly, the sum of the roots for the second equation is -b/a. Since 'p' is a common root, the other root of the second equation is -b/a - p.

Given that the coefficients of x², x, and the constant term in the two equations are equal (since they have the same roots), we can equate the coefficients to find the values of 'a' and 'b'.

From the coefficient of x², we get a = 1.

From the coefficient of x, we get b = -1 - p = -1 - (-b/a - p) = -1 - (-(-1)/1 - p) = -1 - (1 - p) = -p.

Therefore, a + b = 1 - p.

Since 'p' is a common root of the first equation, it must satisfy the equation x² + x + 2 = 0. Substituting 'p' into this equation gives p² + p + 2 = 0. Solving this quadratic equation for 'p' gives p = -1 ± sqrt(1 - 4*2) / 2 = -1 ± sqrt(-7) / 2. Since 'p' is a real number, the discriminant must be non-negative. Therefore, the only possible value for 'p' is -1.

Substituting p = -1 into the equation a + b = 1 - p gives a + b = 1 - (-1) = 2.

Therefore, the value of (a + b) is 2.