I need help with figuring out how to do these problems to get the right answer. Appreciate any help. Thanks.

Multiple choice:
1. What is the value of n so that the expression x² + 12x + n is a perfect square trinomial?

6
36
72
144

2. What are the solutions of the equation x2 + 4x = 96?

2, –48
–12, 8
–2, 48
12, –8

3. What are the solutions of the equation x2 + 14x = –130?

2, –16
10, –13
–2, 16
no solution

4. A rocket is launched from atop a 105-foot cliff with an initial velocity of 156 ft/s. The height of the rocket above the ground at time t is given by h = –16t2 + 156t + 105. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.

4.9 s
9.8 s
0.6 s
10.4 s

since (x+a)^2 = x^2+2a+a^2, you need 2a=12, and n=a^2

x^2+4x-96=0
what factors of 96 are 4 apart?
(x-12)(x+8) = 0

x^2+14x+130=0
the discriminant is 196-420 < 0

set h=0 and solve for t
–16t^2 + 156t + 105 = 0

Sure! I can help you with these problems. Here's a step-by-step explanation on how to solve each one:

1. To determine the value of n in the expression x² + 12x + n so that it is a perfect square trinomial, we need to use the formula for a perfect square trinomial: (a + b)² = a² + 2ab + b².

In this case, the coefficient of x is 12. To find half of this coefficient, we divide it by 2: 12/2 = 6.
So, the square of 6 is 36. Therefore, the value of n that makes the expression a perfect square trinomial is 36.

2. To find the solutions of the equation x² + 4x = 96, we need to set the equation equal to 0 and solve for x.

First, we subtract 96 from both sides of the equation: x² + 4x - 96 = 0.

Next, we can factor the quadratic equation: (x + 12)(x - 8) = 0.

Setting each factor equal to zero, we get x + 12 = 0 or x - 8 = 0.

Solving for x, we have x = -12 or x = 8. Therefore, the solutions of the equation are x = -12 and x = 8.

3. Similar to the previous problem, we set the equation x² + 14x = -130 equal to zero and solve for x.

Subtracting -130 from both sides of the equation, we have: x² + 14x + 130 = 0.

Factoring the quadratic equation, we get: (x + 10)(x + 13) = 0.

Setting each factor equal to zero, we have x + 10 = 0 or x + 13 = 0.

Solving for x, we find x = -10 or x = -13. Therefore, the solutions of the equation are x = -10 and x = -13.

4. To determine when the rocket will hit the ground, we need to find the value of t when the height, h, is equal to 0.

The height of the rocket at time t is given by the equation h = -16t² + 156t + 105.

Setting h equal to 0, we get: -16t² + 156t + 105 = 0.

In order to solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a.

For this equation, a = -16, b = 156, and c = 105.

Plugging these values into the quadratic formula and solving for t, we find two solutions: t = 0.6 seconds and t = 10.4 seconds.

Therefore, the rocket will hit the ground approximately 0.6 seconds after it is launched.