Question

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

Answer 1

Answer:

-5 and -9/2

Step-by-step explanation:

Just did it

Answer 2

Answer:

-5

negative nine-halves

Step-by-step explanation:

we know that

In the quadratic equation $ax^{2} +bx+c=0$

If $b^{2}-4ac < 0$

then

The system has no real numbers solutions

we have

$-x^{2} +3x+c=0$

so

$a=-1,b=3$

substitute

$3^{2}-4(-1)c < 0$

$9+4c < 0$

$c < -\frac{9}{4}$

Verify each case

case 1) -5

For c=-5

substitute

$-5 < -\frac{9}{4}$

$-20 < 9$ ------> 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

$-9/2 < -\frac{9}{4}$

$-36 < -18$ ------> 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

$-1/4 < -\frac{9}{4}$

$-4 < -36$ ------> 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

$1 < -\frac{9}{4}$ ------> 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

$9/4< -\frac{9}{4}$ ------> is not true

therefore

The value of c=9/4 will not cause the quadratic equation  to have no real number solutions