GED Math : Algebra

Study concepts, example questions & explanations for GED Math

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Example Questions

Example Question #2 : Simplifying Quadratics

Add:

Possible Answers:

Correct answer:

Explanation:

can be determined by adding the coefficients of like terms. We can do this vertically as follows:

Example Question #72 : Quadratic Equations

Which of the following expressions is equivalent to the product?

Possible Answers:

Correct answer:

Explanation:

Use the difference of squares pattern

with  and  :

Example Question #1 : Simplifying Quadratics

Which of the following expressions is equivalent to the product?

Possible Answers:

Correct answer:

Explanation:

Use the difference of squares pattern

with  and  :

Example Question #2 : Simplifying Quadratics

Simplify:

Possible Answers:

Correct answer:

Explanation:

Start by factoring the numerator. Notice that each term in the numerator has an , so we can write the following:

Next, factor the terms in the parentheses. You will want two numbers that multiply to  and add to .

Next, factor the denominator. For the denominator, we will want two numbers that multiply to  and add to .

Now that both the denominator and numerator have been factored, rewrite the fraction in its factored form.

Cancel out any terms that appear in both the numerator and denominator.

Example Question #7 : Simplifying Quadratics

Simplify the following expression:

Possible Answers:

Correct answer:

Explanation:

Start by factoring the numerator.

To factor the numerator, you will need to find  numbers that add up to  and multiply to .

Next, factor the denominator.

To factor the denominator, you will need to find two numbers that add up to  and multiply to .

Rewrite the fraction in its factored form.

Since  is found in both numerator and denominator, they will cancel out.

Example Question #1 : Simplifying Quadratics

Simplify:

Possible Answers:

Correct answer:

Explanation:

We need to factor both the numerator and the denominator to determine what can cancel each other out.

If we factor the numerator:

Two numbers which add to 6 and multiply to give you -7.

Those numbers are 7 and -1.

If we factor the denominator:

First factor out a 2  

Two numbers which add to -4 and multiply to give you 3

Those numbers are -3 and -1

Now we can re-write our expression with a product of factors:

We can divide  and  to give us 1, so we are left with

 

 

 

 

Example Question #1 : Solving By Other Methods

Solve for  by completing the square:

Possible Answers:

Correct answer:

Explanation:

To complete the square, we have to add a number that makes the left side of the equation a perfect square. Perfect squares have the formula .

In this case, .

Add this to both sides:

 

 

 

Example Question #2 : Solving By Other Methods

Solve for :

Possible Answers:

Correct answer:

Explanation:

 can be demonstrated to be a perfect square polynomial as follows:

It can therefore be factored using the pattern

with .

We can rewrite and solve the equation accordingly:

This is the only solution.

Example Question #3 : Solving By Other Methods

Solve for :

Possible Answers:

 or 

 or 

 or 

 or 

Correct answer:

 or 

Explanation:

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form . This equation is not in this form, so we must get it in this form as follows:

We factor the quadratic expression as 

so that  and .

By trial and error, we find that 

, so the equation becomes

Set each linear binomial to 0 and solve separately:

 

 

 

The solution set is .

Example Question #4 : Solving By Other Methods

Solve for :

Possible Answers:

 or 

 or 

 or 

Correct answer:

 or 

Explanation:

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form . This equation is not in this form, so we must get it in this form as follows:

We factor the quadratic expression as 

so that  and .

By trial and error, we find that 

, so the equation becomes

.

Set each linear binomial to 0 and solve separately:

 

 

 

The solutions set is 

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