Equations - Algebra 2

First factor the expression by pulling out :

Factor the expression in parentheses by recognizing that it is a difference of squares:

Set each term equal to 0 and solve for the x values:

(1)

When solving a radical equation the fist step is always to isolate the radical. Subtracting from both sides of equation (1).

Square both sides and expand the right side,

Collect all like-terms on onto one side of the equation and use the quadratic formula to find the roots:

(2)                                                                                                                            Reminder

Recall the general solution for the quadratic equation,

(3)                                                                                                                            Use equation (3) to write the solutions to equation (2) and simplify:

Therefore, the roots to the quadratic equation (2) are:

These represent two possible solutions to equation (1). We must check both of them. This is because one of the steps in solving the original equation involved a squaring operation, which can produce fictitious solutions.

Therefore, the only solution for the for equation (1) is:

(1)

When solving a radical equation the fist step is always to isolate the radical. Subtracting from both sides of equation (1).

Square both sides and expand the right side,

Collect all like-terms on onto one side of the equation and use the quadratic formula to find the roots:

(2)                                                                                                                            Reminder

Recall the general solution for the quadratic equation,

(3)                                                                                                                            Use equation (3) to write the solutions to equation (2) and simplify:

Therefore, the roots to the quadratic equation (2) are:

These represent two possible solutions to equation (1). We must check both of them. This is because one of the steps in solving the original equation involved a squaring operation, which can produce fictitious solutions.

Therefore, the only solution for the for equation (1) is:

First factor the expression by pulling out :

Factor the expression in parentheses by recognizing that it is a difference of squares:

Set each term equal to 0 and solve for the x values:

First factor the expression by pulling out :

Factor the expression in parentheses by recognizing that it is a difference of squares:

Set each term equal to 0 and solve for the x values:

1. 
2. 

Substitute equation 2. into equation 1.,

so,

Substitute into equation 2:

so, the solution is .

1. 
2. 

Substitute equation 2. into equation 1.,

so,

Substitute into equation 2:

so, the solution is .

1. 
2. 

Substitute equation 2. into equation 1.,

so,

Substitute into equation 2:

so, the solution is .

1. 
2. 

Substitute equation 2. into equation 1.,

so,

Substitute into equation 2:

so, the solution is .

(1)

When solving a radical equation the fist step is always to isolate the radical. Subtracting from both sides of equation (1).

Square both sides and expand the right side,

Collect all like-terms on onto one side of the equation and use the quadratic formula to find the roots:

(2)                                                                                                                            Reminder

Recall the general solution for the quadratic equation,

(3)                                                                                                                            Use equation (3) to write the solutions to equation (2) and simplify:

Therefore, the roots to the quadratic equation (2) are:

These represent two possible solutions to equation (1). We must check both of them. This is because one of the steps in solving the original equation involved a squaring operation, which can produce fictitious solutions.

Therefore, the only solution for the for equation (1) is:

(1)

When solving a radical equation the fist step is always to isolate the radical. Subtracting from both sides of equation (1).

Square both sides and expand the right side,

Collect all like-terms on onto one side of the equation and use the quadratic formula to find the roots:

(2)                                                                                                                            Reminder

Recall the general solution for the quadratic equation,

(3)                                                                                                                            Use equation (3) to write the solutions to equation (2) and simplify:

Therefore, the roots to the quadratic equation (2) are:

These represent two possible solutions to equation (1). We must check both of them. This is because one of the steps in solving the original equation involved a squaring operation, which can produce fictitious solutions.

Therefore, the only solution for the for equation (1) is:

Step 1: Multiply first equation by −2 and add the result to the second equation. The result is:

Step 2: Multiply first equation by −3 and add the result to the third equation. The result is:

Step 3: Multiply second equation by −23 and add the result to the third equation. The result is:

Step 4: solve for z.

Step 5: solve for y.

Step 6: solve for x by substituting y=2 and z=1 into the first equation.

Step 1: Multiply first equation by −2 and add the result to the second equation. The result is:

Step 2: Multiply first equation by −3 and add the result to the third equation. The result is:

Step 3: Multiply second equation by −23 and add the result to the third equation. The result is:

Step 4: solve for z.

Step 5: solve for y.

Step 6: solve for x by substituting y=2 and z=1 into the first equation.

Step 1: Multiply first equation by −2 and add the result to the second equation. The result is:

Step 2: Multiply first equation by −3 and add the result to the third equation. The result is:

Step 3: Multiply second equation by −23 and add the result to the third equation. The result is:

Step 4: solve for z.

Step 5: solve for y.

Step 6: solve for x by substituting y=2 and z=1 into the first equation.

Step 1: Multiply first equation by −2 and add the result to the second equation. The result is:

Step 2: Multiply first equation by −3 and add the result to the third equation. The result is:

Step 3: Multiply second equation by −23 and add the result to the third equation. The result is:

Step 4: solve for z.

Step 5: solve for y.

Step 6: solve for x by substituting y=2 and z=1 into the first equation.

Add 9 to both sides:

Divide both sides by 27:

The statement, 'The monthly costly to insure your cars varies directly with the number of cars you own' can be mathematically expressed as . M is the monthly cost, C is the number of cars owned, and k is the constant of variation.

Given that it costs $420 a month to insure 3 cars, we can find the k-value.

Divide both sides by 3.

Now, we have the mathematical relationship.

Finding how much it costs to insure 5 cars can be found by substituting 5 for C and solving for M.

This is a FOIL problem. First, we must set up the problem in a form we can use to create the function. To do this we take the opposite sign of each of the numbers and place them in this format: .

Now we can FOIL.

First:

Outside:

Inside:

Last:

Then add them together to get .

Combine like terms to find the answer, which is .

Tom paints for a total of hours (2 on his own, 2 with Huck's help). Since he paints at a rate of feet per hour, use the formula

(or )

to determine the total length of the fence Tom paints.

feet

Subtracting this from the total length of the fence feet gives the length of the fence Tom will NOT paint: feet. If Huck finishes the job, he will paint that feet of the fence. Using , we can determine how long this will take Huck to do:

hours.

If Huck works hours and Tom works hours, he works more hours than Tom.