you need the discriminant to be negative. So,
16-4ac < 0
Make it easy, with a=1. That means 4c > 16
So try c=5
x^2+4x+5 = 0
16-4ac < 0
Make it easy, with a=1. That means 4c > 16
So try c=5
x^2+4x+5 = 0
Because it had too many imaginary problems.
Now, in order for the quadratic equation ax^2+4x+c=0 to have two imaginary solutions, the discriminant (b^2 - 4ac) must be negative. Since b is 4, let's choose a value for a and find the corresponding value for c that satisfies this condition.
Let's say a = 1, then the equation becomes x^2 + 4x + c = 0. For the discriminant to be negative, we have:
(4)^2 - 4(1)(c) < 0
16 - 4c < 0
-4c < -16
c > 4
So, choosing any positive integer value for c greater than 4, let's say c = 5, the equation becomes:
x^2 + 4x + 5 = 0
Now, this equation has two imaginary solutions.
In this case, b = 4, so the discriminant becomes Δ = 4^2 - 4ac = 16 - 4ac. To have a negative discriminant, we need 16 - 4ac < 0.
We can choose any arbitrary integer value for 'a' and solve for 'c' using the inequality:
16 - 4ac < 0
4ac > 16
ac > 4
Let's choose a = 1. Now, divide both sides of the inequality by 'a' (which is 1 in this case):
c > 4
Therefore, any integer value for 'c' greater than 4 would satisfy the condition for the quadratic equation to have two imaginary solutions when a = 1. Let's choose c = 5 as an example.
The equation ax^2 + 4x + c becomes:
x^2 + 4x + 5 = 0
So, a possible pair of integer values for a and c is a = 1 and c = 5, resulting in the equation x^2 + 4x + 5 = 0.
The discriminant of a quadratic equation in the form ax^2 + bx + c = 0 is given by the formula Δ = b^2 - 4ac, where Δ represents the discriminant.
For the equation to have two imaginary solutions, the discriminant needs to be negative. So, we can set up the inequality Δ < 0.
Substituting the given values into the inequality, we have:
(4^2) - 4(a)(c) < 0
16 - 4ac < 0
Now, let's consider a and c as integers. We need to find a pair of integers that satisfy the inequality.
Let's select a = 1 and c = 4:
16 - 4(1)(4) < 0
16 - 16 < 0
0 < 0
Since 0 is not less than 0, this pair of integer values, a = 1 and c = 4, does not satisfy the condition for two imaginary solutions.
Let's select another pair of integers, a = 4 and c = 1:
16 - 4(4)(1) < 0
16 - 16 < 0
0 < 0
Again, this pair of integer values, a = 4 and c = 1, does not satisfy the condition for two imaginary solutions.
It seems that there are no pairs of integer values for a and c that satisfy the condition of having two imaginary solutions for the given equation ax^2 + 4x + c = 0.