If $\displaystyle a, b \, \varepsilon \, R, \ a\neq 0$ and the quadratic equation $\displaystyle ax^{2}-bx+1=0$ has imaginary roots, then $a + b +1$ is
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positive
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negative
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zero
- depends on the sign of $b$
For ax^2 - bx + 1 = 0 to have imaginary roots, the discriminant D = b^2 - 4a < 0, so b^2 < 4a. Since a != 0 and b^2 >= 0, a must be positive. For x=1, the value is a - b + 1. This is not directly helpful. Let's check the vertex or specific values. If roots are imaginary, the parabola ax^2 - bx + 1 does not cross the x-axis. Since a > 0, the parabola opens upward and is always above the x-axis. Thus, for any x, ax^2 - bx + 1 > 0. At x = -1, a + b + 1 > 0.
Since the quadratic equation has imaginary roots, its discriminant must be negative, meaning (-b)^2 - 4a(1) < 0. This simplifies to b^2 < 4a. Let f(x) = ax^2 - bx + 1, and since a cannot be zero and the roots are non-real, the parabola never crosses the x-axis and maintains a constant sign. Because f(0) = 1 is positive, the parabola must open upwards everywhere, making f(x) positive for all real x. Evaluating the expression at x = 1 gives f(1) = a(1)^2 - b(1) + 1 = a - b + 1, which must be positive; adding 2b to both sides gives a + b + 1 = 2a + (a - b + 1), which is the sum of the positive term (a - b + 1) and the positive term 2a, proving a + b + 1 is positive.