### All GED Math Resources

## Example Questions

### Example Question #1 : Quadratic Equations

Solve the equation by factoring:

**Possible Answers:**

**Correct answer:**

Therefore:

### Example Question #2 : Quadratic Equations

Solve for :

**Possible Answers:**

**Correct answer:**

This is a quadratic equation in standard form, so first we need to factor .

This can be factored out as

where .

By trial and error we find that , so

can be expressed as

.

Set each linear binomial equal to 0 and solve separately:

The solution set is .

### Example Question #3 : Quadratic Equations

Note: Figure NOT drawn to scale.

The above triangular sail has area 600 square feet. What is ?

**Possible Answers:**

**Correct answer:**

The area of a right triangle with legs of length and is

.

Substitute and for and and 600 for , then solve for :

We can now factor the quadratic expression:

Set each linear binomial to 0 and solve to get possible solutions:

Since must be positive, we throw out the negative solution.

.

### Example Question #1 : Solving By Factoring

Solve for :

**Possible Answers:**

**Correct answer:**

This is a factoring problem so we need to get all of the variables on one side and set the equation equal to zero. To do this we subtract from both sides to get

Think of the equation in this format to help with the following explanation.

We must then factor to find the solutions for . To do this we must make a factor tree of which is 28 in this case to find the possible solutions. The possible numbers are , , .

Since is positive we know that our factoring will produce two positive numbers.

We then use addition with the factoring tree to find the numbers that add together to equal . So , , and

Success! 14 plus 2 equals . We then plug our numbers into the factored form of

We know that anything multiplied by 0 is equal to 0 so we plug in the numbers for which make each equation equal to 0 so in this case .

### Example Question #321 : Variables

Write the equation of a circle having (3, 4) as center and a radius of .

**Possible Answers:**

**Correct answer:**

The center is located at (3,4) which means the standard equation of a circle which is:

becomes

which equals to

### Example Question #1 : Solving By Factoring

Simplify:

**Possible Answers:**

**Correct answer:**

Change division into multiplication by the reciprocal which gives us the following

Now

this results in the following:

Simplification gives us

which equals

### Example Question #1 : Quadratic Equations

Factor the following expression.

**Possible Answers:**

**Correct answer:**

This expression involves the difference of two cubic terms. To factor an expression in this format, we can use a special formula.

Before we can use this formula, we need to manipulate our original expression to identify and .

Comparing this with the formula, and . Now we can use the formula to factor.

### Example Question #6 : Quadratic Equations

Factor the following expression.

**Possible Answers:**

Not factorable

**Correct answer:**

This problem involves the difference of two cubic terms. We need to use a special factoring formula that will allow us to factor this equation.

But before we can use this formula, we need to manipulate to make it more similar to the left hand side of the special formula. We do this by making the coefficients (343 and 64) part of the cubic power.

Comparing this with , and .

Plug these into the formula.

### Example Question #7 : Quadratic Equations

Factor:

**Possible Answers:**

**Correct answer:**

Begin by factoring out a 2:

Then, we recognize that the trinomial can be factored into two terms, each beginning with :

Since the last term is negative, the signs of the two terms are going to be opposite (i.e. one positive and one negative):

Finally, we need two numbers whose product is negative thirty-five and whose sum is positive two. The numbers and fit this description. So, the factored trinomial is:

### Example Question #351 : Polynomials

Solve for :

**Possible Answers:**

**Correct answer:**

You can factor this trinomial by breaking it up into two binomials that lead with :

You will fill in the binomials by finding two factors of 36 that add up to 5. This is achieved with positive 9 and negative 4:

You can then set each of the two binomials equal to 0 and solve for :

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