To determine the number of real number solutions for the quadratic equation [tex]\(-4j^2 + 3j - 28 = 0\)[/tex], we can use the discriminant of the quadratic formula.
The standard form of a quadratic equation is [tex]\(ax^2 + bx + c = 0\)[/tex]. Here, the coefficients are:
- [tex]\(a = -4\)[/tex]
- [tex]\(b = 3\)[/tex]
- [tex]\(c = -28\)[/tex]
The discriminant ([tex]\(\Delta\)[/tex]) is given by the formula:
[tex]\[
\Delta = b^2 - 4ac
\][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] into the formula:
[tex]\[
\Delta = (3)^2 - 4(-4)(-28)
\][/tex]
[tex]\[
\Delta = 9 - 4 \cdot (-4) \cdot (-28)
\][/tex]
[tex]\[
\Delta = 9 - 4 \cdot 4 \cdot 28
\][/tex]
[tex]\[
\Delta = 9 - 449
\][/tex]
[tex]\[
\Delta = -439
\][/tex]
Now, we examine the value of the discriminant:
- If [tex]\(\Delta > 0\)[/tex], there are 2 real solutions.
- If [tex]\(\Delta = 0\)[/tex], there is 1 real solution.
- If [tex]\(\Delta < 0\)[/tex], there are no real solutions.
Since [tex]\(\Delta = -439\)[/tex], which is less than 0, there are no real number solutions.
The best answer for the question is:
A. 0
