Find the value(s) of "k" that will cause the equation 4x^2 +kx + 4 to have one real solution.

 k = {8, -8}

Step-by-step explanation:

In order for a quadratic equation to have exactly one solution, the discriminant must equal zero.       →        b² - 4ac = 0

  4x² + kx + 4 = 0

  ↓       ↓      ↓

a=4    b=k    c=4

b² - 4ac = 0

k² - 4(4)(4) = 0

k²             = 64

           k = √64

           k = ± 8

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

the answer is b.

64

step-by-step explanation:

c=2πr=2·π·8≈64