For what real values of a does the quadratic 4x^2 + ax + 25 have nonreal roots? Give your answer as an interval.
when b^2 - 4 a c is negative
a^2 - 400 <0
-20 < a < +20
Please convert that to an interval!
To find the real values of "a" for which the quadratic equation has nonreal (complex) roots, we need to consider the discriminant of the quadratic equation. The discriminant, denoted as ∆, is the expression under the square root sign in the quadratic formula.
The general form of a quadratic equation is given by ax^2 + bx + c = 0, where "a," "b," and "c" are constants.
In this case, the equation is 4x^2 + ax + 25 = 0. Comparing this to the general form, we have a = 4, b = a, and c = 25.
The discriminant, ∆, is calculated as follows:
∆ = b^2 - 4ac
Substituting the values, we get:
∆ = (a)^2 - 4(4)(25)
= a^2 - 16(25)
= a^2 - 400
For the quadratic equation to have nonreal roots, the discriminant must be negative (∆ < 0). Therefore, we can set up the inequality:
a^2 - 400 < 0
Solving this inequality will give us the range of real values for "a." Let's proceed:
a^2 - 400 < 0
a^2 < 400
|a| < √400
|a| < 20
Considering both the positive and negative values of "a," we can write the solution as:
-20 < a < 20
Thus, the real values of "a" for which the quadratic equation 4x^2 + ax + 25 has nonreal roots form the interval (-20, 20).