Answer:
-5
negative nine-halves
Step-by-step explanation:
we know that
In the quadratic equation [tex]ax^{2} +bx+c=0[/tex]
If [tex]b^{2}-4ac < 0[/tex]
then
The system has no real numbers solutions
we have
[tex]-x^{2} +3x+c=0[/tex]
so
[tex]a=-1,b=3[/tex]
substitute
[tex]3^{2}-4(-1)c < 0[/tex]
[tex]9+4c < 0[/tex]
[tex]c < -\frac{9}{4}[/tex]
Verify each case
case 1) -5
For c=-5
substitute
[tex]-5 < -\frac{9}{4}[/tex]
[tex]-20 < 9[/tex] ------> is true
therefore
The value of c=-5 will cause the quadratic equation to have no real number solutions
case 2) negative nine-halves
For c=-9/2
substitute
[tex]-9/2 < -\frac{9}{4}[/tex]
[tex]-36 < -18[/tex] ------> is true
therefore
The value of c=-9/2 will cause the quadratic equation to have no real number solutions
case 3) negative one-quarter
For c=-1/4
substitute
[tex]-1/4 < -\frac{9}{4}[/tex]
[tex]-4 < -36[/tex] ------> is not true
therefore
The value of c=-1/4 will not cause the quadratic equation to have no real number solutions
case 4) 1
For c=1
substitute
[tex]1 < -\frac{9}{4}[/tex] ------> is not true
therefore
The value of c=1 will not cause the quadratic equation to have no real number solutions
case 5) 9 Over 4
For c=9/4
substitute
[tex]9/4< -\frac{9}{4}[/tex] ------> is not true
therefore
The value of c=9/4 will not cause the quadratic equation to have no real number solutions
