Question

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

–5
negative nine-halves
negative one-quarter
1
StartFraction 9 Over 4 EndFraction

Answer 1

-5 and -9/2

Explanation:

Just did it

Answer 2

-5

negative nine-halves

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 < -(9)/(4)`

Verify each case

case 1) -5

For c=-5

substitute

`-5 < -(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 < -(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 < -(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 < -(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< -(9)/(4)` ------> is not true

therefore

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