For what values of b, the equation x^2−ax+b=0 has a solution for any value of a?

To find the values of b for which the equation x^2 - ax + b = 0 has a solution for any value of a, we need to determine the conditions under which the quadratic equation has real roots.

For a quadratic equation, the discriminant can be used to determine the nature of its roots. The discriminant (Δ) is given by the formula Δ = b^2 - 4ac, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

In this case, comparing the given equation x^2 - ax + b = 0 with the general quadratic equation, we can see that a = -a, b = b, and c = 0.

Substituting these values into the discriminant formula, we get Δ = (-a)^2 - 4(1)(b) = a^2 - 4b.

For the quadratic equation to have real roots, the discriminant should be greater than or equal to zero, Δ ≥ 0.

Therefore, we have the inequality a^2 - 4b ≥ 0.

Simplifying this inequality, we obtain b ≤ a^2/4.

Hence, for any value of a, the equation x^2 - ax + b = 0 has a solution when b is less than or equal to a^2/4. In other words, the values of b lie in the interval (-∞, a^2/4].

well, b^2-4(-a)b has to be positive

b^2+4ab>0
b >-4a