Call Now to Set Up Tutoring:

(888) 888-0446

GED Math : Algebra

Study concepts, example questions & explanations for GED Math

Create An Account

All GED Math Resources

4 Diagnostic Tests 263 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #391 : Algebra

Add:

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) + (1.7x^{2} - 3.9x - 3.2)$

Possible Answers:

$4.8x^{2} +8.4x -2.7$

$4.8x^{2} + 0.6x -3.7$

$4.8x^{2} +8.4x -3.7$

$4.8x^{2} + 0.6x -2.7$

Correct answer:

$4.8x^{2} + 0.6x -2.7$

Explanation:

$\left (3.1x^{2}+ 4.5x + 0. 5 \right ) + (1.7x^{2} - 3.9x - 3.2)$

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

$\begin{matrix} \; \; 3.1x^{2}+ 4.5x + 0. 5 \\ + \underline{1.7x^{2} - 3.9x - 3.2 }\\ \; \; 4.8x^{2} + 0.6x -2.7\end{matrix}$

Report an Error

Example Question #392 : Algebra

Which of the following expressions is equivalent to the product?

$\left ( \frac{1}{5x} - 5x \right )\left ( \frac{1}{5x}+5x\right )$

Possible Answers:

$\frac{-625x ^{4 }+50x^{2}-1}{25x^{2}}$

$\frac{625x ^{4 }- 1}{25x^{2}}$

$\frac{625x ^{4 }-50x^{2}+ 1}{25x^{2}}$

$\frac{-625x ^{4} + 1 }{25x^{2}}$

Correct answer:

$\frac{-625x ^{4} + 1 }{25x^{2}}$

Explanation:

Use the difference of squares pattern

$(A + B)(A - B) = A^{2} - B^{2}$

with $A = \frac{1}{5x}$ and B =5x :

$\left ( \frac{1}{5x} - 5x \right )\left ( \frac{1}{5x}+5x\right )$

$=\left ( \frac{1}{5x} \right )^{2} - \left ( 5x\right )^{2}$

$= \frac{1}{25x^{2}} - 25x ^{2}$

$= \frac{1}{25x^{2}} - \frac{25x ^{2} \cdot 25x ^{2} }{25x ^{2} }$

$= \frac{1}{25x^{2}} - \frac{625x ^{4} }{25x ^{2} }$

$= \frac{1-625x ^{4} }{25x^{2}}$

$= \frac{-625x ^{4} + 1 }{25x^{2}}$

Report an Error

Example Question #4 : Simplifying Quadratics

Which of the following expressions is equivalent to the product?

$\left ( \frac{1}{3x} - \frac{2}{3}\right )\left ( \frac{1}{3x} + \frac{2}{3}\right )$

Possible Answers:

$\frac{-4x^{2}+ 1}{9x^{2}}$

$\frac{4x^{2}-4x + 1}{9x^{2}}$

$\frac{ 4x^{2}- 1}{9x^{2}}$

$\frac{-4x^{2}+4x - 1}{9x^{2}}$

Correct answer:

$\frac{-4x^{2}+ 1}{9x^{2}}$

Explanation:

Use the difference of squares pattern

$(A + B)(A - B) = A^{2} - B^{2}$

with $A = \frac{1}{3x}$ and $B = \frac{2}{3}$ :

$\left ( \frac{1}{3x} - \frac{2}{3}\right )\left ( \frac{1}{3x} + \frac{2}{3}\right )$

$=\left ( \frac{1}{3x} \right ) ^{2} -\left ( \frac{2}{3}\right ) ^{2}$

$= \frac{1}{9x^{2}} - \frac{4}{9}$

$= \frac{1}{9x^{2}} - \frac{4x^{2}}{9x^{2}}$

$= \frac{1-4x^{2}}{9x^{2}}$

$= \frac{-4x^{2}+ 1}{9x^{2}}$

Report an Error

Example Question #6 : Simplifying Quadratics

Simplify:

$\frac{x^3+7x^2+10x}{x^2+3x+2}$

Possible Answers:

$\frac{x+1}{x+5}$

$\frac{x(x+5)}{x+1}$

$\frac{x(x+2)}{x+1}$

$\frac{(x+1)(x+5)}{x}$

Correct answer:

$\frac{x(x+5)}{x+1}$

Explanation:

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

x^3+7x^2+10x=x(x^2+7x+10)

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

x(x^2+7x+10)=x(x+5)(x+2)

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

x^2+3x+2=(x+2)(x+1)

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

$\frac{x^3+7x^2+10x}{x^2+3x+2}=\frac{x(x+5)(x+2)}{(x+2)(x+1)}$

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

$\frac{x(x+5)(x+2)}{(x+2)(x+1)}=\frac{x(x+5)}{x+1}$

Report an Error

Example Question #6 : Simplifying Quadratics

Simplify the following expression:

$\frac{x^2+5x+6}{x^2+3x+2}$

Possible Answers:

$\frac{x+2}{x+1}$

x+2

$\frac{x+1}{x+3}$

$\frac{x+3}{x+1}$

Correct answer:

$\frac{x+3}{x+1}$

Explanation:

Start by factoring the numerator.

x^2+5x+6

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

x^2+5x+6=(x+2)(x+3)

Next, factor the denominator.

x^2+3x+2

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

x^2+3x+2=(x+1)(x+2)

Rewrite the fraction in its factored form.

$\frac{x^2+5x+6}{x^2+3x+2}=\frac{(x+2)(x+3)}{(x+1)(x+2)}$

Since x+2 is found in both numerator and denominator, they will cancel out.

$\frac{(x+2)(x+3)}{(x+1)(x+2)}=\frac{x+3}{x+1}$

Report an Error

Example Question #5 : Simplifying Quadratics

Simplify:

$\frac{x^{2}+6x-7}{2x^{2}-8x+6}$

Possible Answers:

$\frac{(x-7)(x+1)}{2(x-1)(x-3)}$

$\frac{x+7}{2x+6}$

$\frac{x+7}{2(x-3)}$

$\frac{x+7}{x-3}$

$\frac{(x-1)(x+7)}{(2x-1)(x+3)}$

Correct answer:

$\frac{x+7}{2(x-3)}$

Explanation:

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

If we factor the numerator:

$x^{2}+6x-7$

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

Those numbers are 7 and -1.

$x^{2}+6x-7=\textbf{(x+7)(x-1)}$

If we factor the denominator:

$2x^{2}-8x+6$

First factor out a 2 $\rightarrow$ $2(x^{2}-4x+3)$

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

Those numbers are -3 and -1

$2x^{2}-8x+6=\textbf{2(x-1)(x-3)}$

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

$\frac{(x+7)(x-1)}{2(x-1)(x-3)}$

We can divide (x-1) and (x-1) to give us 1, so we are left with

$\frac{(x+7)}{2(x-3)}$

Report an Error

Example Question #393 : Algebra

Solve for by completing the square:

$\small 4x^2+12x+2=0$

Possible Answers:

$\small x=\frac{-3\pm \sqrt{-11}}{2}$

$\small x=\frac{-3\pm \sqrt7}{2}$

$\small x=\frac{-9\pm \sqrt{-11}}{2}$

$\small x=\frac{-9\pm \sqrt7}{2}$

Correct answer:

$\small x=\frac{-3\pm \sqrt7}{2}$

Explanation:

$\small 4x^2+12x+2=0$

$\small 4x^2+12+2-2=0-2$

$\small 4x^2+12x=-2$

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 $\small \small \small a^2+2ab+b^2$ .

In this case, $\small \small a=2x,\ 2ab=12x$ .

$\small \small 12=2(2x)(b)=4xb; b=3$

$\small b^2=3^2=9$

Add this to both sides:

$\small 4x^2+12x+9=-2+9$

$\small \small 4x^2+12x+9=7$

$\small (2x+3)^2=7$

$\small \sqrt{(2x+3)^2}=\pm \sqrt7$

$\small 2x+3=\pm \sqrt{7}$

$\small 2x+3-3=-3\pm \sqrt7$

$\small 2x=-3\pm \sqrt7$

$\small \frac{2x}{2}=\frac{-3\pm \sqrt7}{2}$

$\small x=\frac{-3\pm \sqrt7}{2}$

Report an Error

Example Question #394 : Algebra

Solve for :

$4x ^{2} - 20x + 25 = 0$

Possible Answers:

$x = - 2\frac{1}{2} \textrm{ or }x = 2\frac{1}{2}$

$x = \frac{2}{5}$

$x = -\frac{2}{5} \textrm{ or }x = \frac{2}{5}$

$x = 2\frac{1}{2}$

Correct answer:

$x = 2\frac{1}{2}$

Explanation:

$4x ^{2} - 20x + 25$ can be demonstrated to be a perfect square polynomial as follows:

$4x ^{2} - 20x + 25 = \left ( 2x\right )^{2} - 2 (2x)(5) + 5^{2}$

It can therefore be factored using the pattern

$A^{2}-2AB + B^{2} = \left (A - B \right )^{2}$

with A = 2x, B = 5 .

We can rewrite and solve the equation accordingly:

$4x ^{2} - 20x + 25 = 0$

$(2x-5) ^{2} = 0$

2x-5 = 0

2x = 5

$x = 5 \div 2$

$x = 2\frac{1}{2}$

This is the only solution.

Report an Error

Example Question #395 : Algebra

Solve for :

x (x - 7) = 30

Possible Answers:

x = 2 or x = 6

x = -3 or x = 10

x = 9 or x = 13

x = 30 or x = 37

Correct answer:

x = -3 or x = 10

Explanation:

When solving a quadratic equation, it is necessary to write it in standard form first - that is, in the form $ax^{2} + bx + c = 0$ . This equation is not in this form, so we must get it in this form as follows:

x (x - 7) = 30

$x \cdot x -x \cdot 7 = 30$

$x ^{2} -7x = 30$

$x ^{2} -7x - 30 = 0$

We factor the quadratic expression as

(x+M)(x+N)

so that MN = -30 and M + N = -7 .

By trial and error, we find that

M = -10, N = 3 , so the equation becomes

(x-10)(x+3) = 0

Set each linear binomial to 0 and solve separately:

x - 10 = 0

x = 10

x + 3 = 0

x = -3

The solution set is $\left \{ -3, 10 \right \}$ .

Report an Error

Example Question #396 : Algebra

Solve for :

$x^{2} - 12 = 6x+4$

Possible Answers:

x = -4 or x = 4

x = 4

x= -2 or x= 8

x= -8 or x= 2

Correct answer:

x= -2 or x= 8

Explanation:

$x^{2} - 12 = 6x+4$

$x^{2} - 12- 6x - 4 = 6x+4 - 6x - 4$

$x^{2} - 6x - 16 = 0$

We factor the quadratic expression as

(x+M)(x+N)

so that MN = -16 and M + N = -6 .

By trial and error, we find that

M = -8, N = 2 , so the equation becomes

(x-8)(x+2) = 0 .

Set each linear binomial to 0 and solve separately:

x-8=0

x= 8

x+2= 0

x= -2

The solutions set is $\left \{ -2, 8\right \}$

Report an Error

View Tutors

Vicky
Certified Tutor

Clemson University, Doctor of Science, Civil Engineering.

View Tutors

Rachel
Certified Tutor

University of Maryland-College Park, Bachelor of Science, Civil Engineering. Relay Graduate School of Education, Master of Sc...

View Tutors

Charlynn
Certified Tutor

Florida International University, Bachelor of Science, Business Administration and Management.

All GED Math Resources

4 Diagnostic Tests 263 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Statistics Tutors in Phoenix, GMAT Tutors in San Francisco-Bay Area, Chemistry Tutors in New York City, Biology Tutors in Atlanta, ACT Tutors in Boston, French Tutors in Atlanta, Math Tutors in Seattle, Chemistry Tutors in Los Angeles, Calculus Tutors in Los Angeles, Spanish Tutors in New York City

Popular Courses & Classes

MCAT Courses & Classes in Seattle, Spanish Courses & Classes in New York City, MCAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Atlanta, SSAT Courses & Classes in San Diego, GMAT Courses & Classes in New York City, LSAT Courses & Classes in New York City, ISEE Courses & Classes in Boston, ACT Courses & Classes in Phoenix, ISEE Courses & Classes in Los Angeles

Popular Test Prep

GMAT Test Prep in Phoenix, GMAT Test Prep in New York City, LSAT Test Prep in Los Angeles, MCAT Test Prep in Houston, ACT Test Prep in Chicago, ISEE Test Prep in Boston, SSAT Test Prep in Phoenix, LSAT Test Prep in San Diego, LSAT Test Prep in Denver, LSAT Test Prep in Phoenix

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

GED Math : Algebra

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #391 : Algebra

Example Question #392 : Algebra

Example Question #4 : Simplifying Quadratics

Example Question #6 : Simplifying Quadratics

Example Question #6 : Simplifying Quadratics

Example Question #5 : Simplifying Quadratics

Example Question #393 : Algebra

Example Question #394 : Algebra

Example Question #395 : Algebra

Example Question #396 : Algebra

All GED Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: