# I'm in Algebra 1 this year and I really need help solving these equations.

Completing the square

1)c^2-7c= -12
2)x^2-2x-8= 0

1) 2x^2-2 = 16

Any advice will be appreciated!

c^2-7c= -12
What is 1/2 of -7? Square that, and add to both sides.

x^2-2x-8= 0
What is 1/2 of -2? square that, and add to both sides

A cylinder has a radius of 5 in. If the volume of the cylinder is 250* pi in. ^3, what is the height of the cylinder

[Hint: The volume of a cylinder is given by V=pi * r&2 *h
Remember that pi = 3.14

## To solve the equation c^2-7c= -12, we will complete the square. Here's how to do it:

Step 1: Move the constant term to the other side:
c^2 - 7c + 12 = 0

Step 2: Find the value that completes the square. Take half of the coefficient of the linear term (-7 in this case), which is -7/2, and square it:
(-7/2)^2 = 49/4

Step 3: Add this value to both sides of the equation:
c^2 - 7c + 49/4 + 12 = 49/4

Simplifying the equation:
c^2 - 7c + 49/4 + 48/4 = 49/4
c^2 - 7c + (49 + 48)/4 = 49/4
c^2 - 7c + 97/4 = 49/4

Step 4: Rewrite the left side as a perfect square trinomial. In this case, we can rewrite it as (c - 7/2)^2:
(c - 7/2)^2 = 49/4

Step 5: Take the square root of both sides to solve for c - 7/2:
√[(c - 7/2)^2] = √(49/4)
c - 7/2 = ±7/2

Step 6: Solve for c by adding 7/2 to both sides:
c = 7/2 ± 7/2

The solutions are:
c = 7/2 + 7/2 = 7/2 + 7/2 = 7/2 + 7/2 = 7
c = 7/2 - 7/2 = 7/2 - 7/2 = 0

Therefore, the solutions to the equation c^2-7c= -12 are c = 7 and c = 0.

To solve the equation x^2-2x-8= 0, we will also complete the square. Here's how to do it:

Step 1: Move the constant term to the other side:
x^2 - 2x + 8 = 0

Step 2: Find the value that completes the square. Take half of the coefficient of the linear term (-2 in this case), which is -2/2, and square it:
(-2/2)^2 = 1

Step 3: Add this value to both sides of the equation:
x^2 - 2x + 1 + 8 = 1

Simplifying the equation:
x^2 - 2x + 9 = 1

Step 4: Rewrite the left side as a perfect square trinomial. In this case, we can rewrite it as (x - 1)^2:
(x - 1)^2 = 1

Step 5: Take the square root of both sides to solve for x - 1:
√[(x - 1)^2] = √1
x - 1 = ±1

Step 6: Solve for x by adding 1 to both sides:
x = 1 ± 1

The solutions are:
x = 1 + 1 = 2
x = 1 - 1 = 0

Therefore, the solutions to the equation x^2-2x-8= 0 are x = 2 and x = 0.

To solve the incomplete quadratic equation 2x^2-2 = 16, we will complete the square. Here's how to do it:

Step 1: Move the constant term to the other side:
2x^2 - 2 - 16 = 0

Step 2: Combine like terms:
2x^2 - 18 = 0

Step 3: Divide both sides by the leading coefficient (2 in this case):
(2x^2 - 18)/2 = 0/2
x^2 - 9 = 0

Step 4: Find the value that completes the square. Take half of the coefficient of the linear term (0 in this case), which is 0/2 = 0, and square it:
(0)^2 = 0

Step 5: Add this value to both sides of the equation:
x^2 - 9 + 0 = 0

Simplifying the equation:
x^2 - 9 = 0

Step 6: Rewrite the left side as a perfect square trinomial. In this case, we can rewrite it as (x - 3)(x + 3) = 0

Step 7: Solve for x by setting each factor equal to zero:
x - 3 = 0 or x + 3 = 0

Solving for x in each case:
For x - 3 = 0, add 3 to both sides: x = 3
For x + 3 = 0, subtract 3 from both sides: x = -3

Therefore, the solutions to the equation 2x^2-2 = 16 are x = 3 and x = -3.

For the volume of a cylinder problem, we can use the formula V = π * r^2 * h, where V is the volume, r is the radius, and h is the height.

Given that the radius of the cylinder is 5 in and the volume is 250π in^3, we can substitute these values into the formula:

250π = π * (5^2) * h

Simplifying the equation:
250π = 25πh

Divide both sides of the equation by 25π:
250π / (25π) = h

Simplifying the fraction:
10 = h

Therefore, the height of the cylinder is 10 inches.