What values of b will cause 4x^2+bx+25=0 to have one real solution?

The values of b will cause 4x^2+bx+25=0 to have one real solution is -20 and 20

Given the quadratic equation , for this expression to have a real solution, the discriminant must be greater than zero

D > 0

b^2 - 4ac > 0

Given that a = 4, b = b, c = 25

b^2 - 4(4)(25) > 0

b^2 - 400 > 0

b^2 > 400

b > ±20

Hence the values of b will cause 4x^2+bx+25=0 to have one real solution is -20 and 20

Learn more on discriminant here: link

b=20 or b=-20

Step-by-step explanation:

Keep in mind the meanings of the values of the discriminant:

If b^2-4ac=0, then the quadratic will have only 1 solution (double root)

If b^2-4ac<0, then the quadratic will have no real solutions

If b^2-4ac>0, then the quadratic will have 2 unique solutions

In this case, to get one real solution, the discriminant must be set up as b^2-4ac=0. Setting up the equation we get:

b^2-4(4)(25)=0

b^2-400=0

b^2=400

b=20 or b=-20

So the values of b would have to be either 20 or -20

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x