Which values of c will cause the quadratic equation –x2 + 3x + c = 0 to have no real number solutions? Check all that apply.

Given the quadratic function:

-x^2 + 3x + c

a = -1

b = 3

Determine the value of c to which the quadratic function will not have any solution.

Explanation:

The value of c must be a negative number that when multiplied by 4, the result must be greater than the square of 3, therefore, c must be -3 up until - infinity. This can be clearly understood by looking at the following statements based from the quadratic formula.

The Quadratic formula:

x = −b ± √(b^2 − 4ac)/2a

When b^2−4ac=0 there is one real root.

When b^2−4ac>0 there are two real roots.

When b^2−4ac<0 there are no real roots, only a complex number.

This means that b^2 -4ac or the equation inside the square root must equal to a negative number in order to have no real solutions. Negative numbers do not have a square root since it will result to a complex number, hence in our case we have a negative 1, which is the value of a, the in the solution we have -4 multiplied to -a, thus the result will be positive, therefore c must be negative to make the value negative again. In the case of b^2, since we have 3, if we square 3, then the answer will be 9. This means that our value must be greater than 9, and these values include from -3 to - infinity.

Final answer:

c = ( -3, -∞)