How many real solutions does the equation x^2 + bx + 4 = 0 have if b>0

a. none
b. one
c. two
d. cannot be determined

Please help me out! I don't know how to solve this problem, so an explanation would be appreciated; it will also be great if you could supply me with an answer so that I can check to see if the answer I got is correct :) Thank you so much <3 :3

It's real if the discriminant is greater than or equal to 0.

D >= 0
b² - 4ac >= 0
b² - 16 >= 0
b² >= 16
b <= -4 or b >= 4
Infinitely many.

The answer is (D) cannot be determined.

The discriminant of x^2 + bx + 4 = 0 is b^2 - 4(1)(4) = b^2 - 16, which is negative for 0 < b < 4, zero for b = 4, and positive for b > 4. In other words, no real solutions exist for 0 < b < 4, one solution exists for b = 4, and two distinct real solutions exist for b > 4. As you can see, zero, one, or two real solutions can exist depending on the value of b, so we cannot determine how many solutions the equation has.

I hope this helps!

The b^2 - 4ac part of te quadratic formula tells you if roots are imaginary or not.

In this case, 4ac is always 16, so unless b^2 >= 16 the roots are complex.

However, you can't tell how any real solutions the equation has, although you can can say b must be less than or equal to -4 and greater than or equal to positive 4.

D is the correct answer.

Answer is d. cannot be determined.

Reason.
1. When 4>b>0, roots will be imaginary.
2. When b=4, only one root is possible.
3. When b>4, rational roots are possible.

b^2-4ac in this case becomes
b^2-16
If b^2-16 > 0 we have 2 real solutions
If b^2-16 = 0 we have 1 real solution
If b^2-16 < we have 0 real solutions
All we know is that b>0 so it cannot be determined.
Answer d.

d. cannot be determined

If b = 4 there is one solution
If b < 4 there are no solutions
if b > 4 there are two solutions

It has none until b >= 4, so it depends how far greater than zero b is.