We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

Create An Account

All Algebra II Resources

10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #123 : Solving Quadratic Equations

Solve the quadratic equation by completing the square:

$\displaystyle x^2+22x+100=0$

Possible Answers:

$x=-9.01\text{ or }-18.23$ $\displaystyle x=-9.01\text{ or }-18.23$

$x=-9.29\text{ or }-21.32$ $\displaystyle x=-9.29\text{ or }-21.32$

$x=-5.45\text{ or }-15.61$ $\displaystyle x=-5.45\text{ or }-15.61$

$x=-6.42\text{ or }-15.58$ $\displaystyle x=-6.42\text{ or }-15.58$

Correct answer:

$x=-6.42\text{ or }-15.58$ $\displaystyle x=-6.42\text{ or }-15.58$

Explanation:

Start by moving the number to the right of the equation so that all the terms with $\displaystyle x$ values are alone:

$\displaystyle x^2+22x+100=0$

$\displaystyle x^2+22x=-100$

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the $\displaystyle x$ term, divide it by $\displaystyle 2$, then square it:

Coefficient: $\displaystyle 22$

$\text{Term to add}=(\frac{22}{2})^2=(11)^2=121$ $\displaystyle \text{Term to add}=(\frac{22}{2})^2=(11)^2=121$

Add this number to both sides of the equation:

$\displaystyle x^2+22x+121=-100+121=21$

Simplify the left side of the equation.

$\displaystyle (x+11)^2=21$

Now solve for $\displaystyle x$.

$x+11=\pm\sqrt{21}$ $\displaystyle x+11=\pm\sqrt{21}$

$x=-11+\sqrt{21}=-6.42$ $\displaystyle x=-11+\sqrt{21}=-6.42$, or

$x=-11-\sqrt{21}=-15.58$ $\displaystyle x=-11-\sqrt{21}=-15.58$

Make sure to round to $\displaystyle 2$ places after the decimal.

Report an Error

Example Question #55 : Completing The Square

Solve the quadratic equation by completing the square:

$\displaystyle x^2-30x+98=0$

Possible Answers:

$x=21.92\text{ or }12.28$ $\displaystyle x=21.92\text{ or }12.28$

$x=26.27\text{ or }3.73$ $\displaystyle x=26.27\text{ or }3.73$

$x=29.71\text{ or }9.22$ $\displaystyle x=29.71\text{ or }9.22$

$x=25.67\text{ or }8.21$ $\displaystyle x=25.67\text{ or }8.21$

Correct answer:

$x=26.27\text{ or }3.73$ $\displaystyle x=26.27\text{ or }3.73$

Explanation:

Start by moving the number to the right of the equation so that all the terms with $\displaystyle x$ values are alone:

$\displaystyle x^2-30x+98=0$

$\displaystyle x^2-30x=-98$

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the $\displaystyle x$ term, divide it by $\displaystyle 2$, then square it:

Coefficient: $\displaystyle 30$

$\text{Term to add}=(\frac{30}{2})^2=(15)^2=225$ $\displaystyle \text{Term to add}=(\frac{30}{2})^2=(15)^2=225$

Add this number to both sides of the equation:

$\displaystyle x^2-30x+225=-98+225=127$

Simplify the left side of the equation.

$\displaystyle (x-15)^2=127$

Now solve for $\displaystyle x$.

$x-15=\pm\sqrt{127}$ $\displaystyle x-15=\pm\sqrt{127}$

$x=15+\sqrt{127}=26.27$ $\displaystyle x=15+\sqrt{127}=26.27$, or

$x=15-\sqrt{127}=3.73$ $\displaystyle x=15-\sqrt{127}=3.73$

Make sure to round to $\displaystyle 2$ places after the decimal.

Report an Error

Example Question #56 : Completing The Square

Solve the following quadratic equation by completing the square:

$\displaystyle x^2-16x+40=0$

Possible Answers:

$x=-5.19\text{ or }10.72$ $\displaystyle x=-5.19\text{ or }10.72$

$x=4.19\text{ or }19.52$ $\displaystyle x=4.19\text{ or }19.52$

$x=-3.09\text{ or }14.98$ $\displaystyle x=-3.09\text{ or }14.98$

$x=3.10\text{ or }12.90$ $\displaystyle x=3.10\text{ or }12.90$

Correct answer:

$x=3.10\text{ or }12.90$ $\displaystyle x=3.10\text{ or }12.90$

Explanation:

Start by moving the number to the right of the equation so that all the terms with $\displaystyle x$ values are alone:

$\displaystyle x^2-16x+40=0$

$\displaystyle x^2-16x=-40$

Now, you need to figure out what number to add to both sides. To do so, take the coefficient in front of the $\displaystyle x$ term, divide it by $\displaystyle 2$, then square it:

Coefficient: $\displaystyle -16$

$\text{Term to add}=(\frac{-16}{2})^2=(-8)^2=64$ $\displaystyle \text{Term to add}=(\frac{-16}{2})^2=(-8)^2=64$

Add this number to both sides of the equation:

$\displaystyle x^2-16x+64=-40+64=24$

Simplify the left side of the equation.

$\displaystyle (x-8)^2=24$

Now solve for $\displaystyle x$.

$x-8=\pm\sqrt{24}$ $\displaystyle x-8=\pm\sqrt{24}$

$x=8+\sqrt{24}=12.90$ $\displaystyle x=8+\sqrt{24}=12.90$, or

$x=8-\sqrt{24}=3.10$ $\displaystyle x=8-\sqrt{24}=3.10$

Make sure to round to $\displaystyle 2$ places after the decimal.

Report an Error

Example Question #51 : Completing The Square

Which of the following is the same after completing the square?

$\displaystyle 3x^2+x-6 =0$

Possible Answers:

$(x+\frac{1}{6})^2 =\frac{73}{36}$ $\displaystyle (x+\frac{1}{6})^2 =\frac{73}{36}$

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

$(x-\frac{1}{3})^2 =\frac{7}{2}$ $\displaystyle (x-\frac{1}{3})^2 =\frac{7}{2}$

$(x-\frac{1}{6})^2 =\frac{73}{36}$ $\displaystyle (x-\frac{1}{6})^2 =\frac{73}{36}$

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

Correct answer:

$(x+\frac{1}{6})^2 =\frac{73}{36}$ $\displaystyle (x+\frac{1}{6})^2 =\frac{73}{36}$

Explanation:

Divide by three on both sides.

$\frac{3x^2+x-6 }{3}=\frac{0}{3}$ $\displaystyle \frac{3x^2+x-6 }{3}=\frac{0}{3}$

$x^2+\frac{1}{3}x-2 = 0$ $\displaystyle x^2+\frac{1}{3}x-2 = 0$

Add two on both sides.

$x^2+\frac{1}{3}x-2 +2= 0+2$ $\displaystyle x^2+\frac{1}{3}x-2 +2= 0+2$

$x^2+\frac{1}{3}x = 2$ $\displaystyle x^2+\frac{1}{3}x = 2$

To complete the square, we will need to divide the one-third coefficient by two, which is similar to multiplying by one half, square the quantity, and add the two values on both sides.

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

Simplify both sides.

$x^2+\frac{1}{3}x+ \frac{1}{36} = 2+ \frac{1}{36}$ $\displaystyle x^2+\frac{1}{3}x+ \frac{1}{36} = 2+ \frac{1}{36}$

Factor the left side, and combine the terms on the right.

$(x+\frac{1}{6})^2 = \frac{72}{36}+ \frac{1}{36}$ $\displaystyle (x+\frac{1}{6})^2 = \frac{72}{36}+ \frac{1}{36}$

The answer is: $(x+\frac{1}{6})^2 =\frac{73}{36}$ $\displaystyle (x+\frac{1}{6})^2 =\frac{73}{36}$

Report an Error

Example Question #1581 : Algebra Ii

Solve for $\displaystyle x$ by completing the square.

$\displaystyle x^2+8x-2=0$

Possible Answers:

$x=1.24\text{ and }x=-10.24$ $\displaystyle x=1.24\text{ and }x=-10.24$

$x=0.39\text{ and }x=-4.39$ $\displaystyle x=0.39\text{ and }x=-4.39$

$x=0.24\text{ and }x=-8.24$ $\displaystyle x=0.24\text{ and }x=-8.24$

$x=0.97\text{ and }x=-8.97$ $\displaystyle x=0.97\text{ and }x=-8.97$

Correct answer:

$x=0.24\text{ and }x=-8.24$ $\displaystyle x=0.24\text{ and }x=-8.24$

Explanation:

$\displaystyle x^2+8x-2=0$

Start by adding $\displaystyle 2$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle x^2+8x=2$

Now, look at the coefficient of the $\displaystyle x$-term. To complete the square, divide this coefficient by $\displaystyle 2$, then square the result. Add this term to both sides of the equation.

$\displaystyle x^2+8x+16=2+16$

Rewrite the left side of the equation in the squared form.

$\displaystyle (x+4)^2=18$

Take the square root of both sides.

$x+4=\pm \sqrt{18}$ $\displaystyle x+4=\pm \sqrt{18}$

Now solve for $\displaystyle x$.

$x=-4+\sqrt{18}=0.24$ $\displaystyle x=-4+\sqrt{18}=0.24$

$x=-4-\sqrt{18}=-8.24$ $\displaystyle x=-4-\sqrt{18}=-8.24$

Round to two places after the decimal.

Report an Error

Example Question #1582 : Algebra Ii

Solve for $\displaystyle x$ by completing the square.

$\displaystyle x^2-6x+2=0$

Possible Answers:

$x=5.65\text{ and }x=0.35$ $\displaystyle x=5.65\text{ and }x=0.35$

$x=1.95\text{ and }x=5.05$ $\displaystyle x=1.95\text{ and }x=5.05$

$x=-8.75\text{ and }x=-2.25$ $\displaystyle x=-8.75\text{ and }x=-2.25$

$x=10.55\text{ and }x=2.45$ $\displaystyle x=10.55\text{ and }x=2.45$

Correct answer:

$x=5.65\text{ and }x=0.35$ $\displaystyle x=5.65\text{ and }x=0.35$

Explanation:

$\displaystyle x^2-6x+2=0$

Start by subtracting $\displaystyle 2$ from both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle x^2-6x=-2$

$\displaystyle x^2-6x+9=-2+9$

Rewrite the left side of the equation in the squared form.

$\displaystyle (x-3)^2=7$

Take the square root of both sides.

$x-3=\pm \sqrt{7}$ $\displaystyle x-3=\pm \sqrt{7}$

Now solve for $\displaystyle x$.

$x=3+\sqrt{7}=5.65$ $\displaystyle x=3+\sqrt{7}=5.65$

$x=3-\sqrt{7}=0.35$ $\displaystyle x=3-\sqrt{7}=0.35$

Round to two places after the decimal.

Report an Error

Example Question #1582 : Algebra Ii

Solve for $\displaystyle x$ by completing the square.

$\displaystyle x^2-10x+20=0$

Possible Answers:

$x=3.33\text{ and }x=8.67$ $\displaystyle x=3.33\text{ and }x=8.67$

$x=-5.21\text{ and }x=-9.79$ $\displaystyle x=-5.21\text{ and }x=-9.79$

$x=8.06\text{ and }x=1.94$ $\displaystyle x=8.06\text{ and }x=1.94$

$x=7.24\text{ and }x=2.76$ $\displaystyle x=7.24\text{ and }x=2.76$

Correct answer:

$x=7.24\text{ and }x=2.76$ $\displaystyle x=7.24\text{ and }x=2.76$

Explanation:

$\displaystyle x^2-10x+20=0$

Start by subtracting $\displaystyle 20$ from both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle x^2-10x=-20$

$\displaystyle x^2-10x+25=-20+25$

Rewrite the left side of the equation in the squared form.

$\displaystyle (x-5)^2=5$

Take the square root of both sides.

$x-5=\pm \sqrt{5}$ $\displaystyle x-5=\pm \sqrt{5}$

Now solve for $\displaystyle x$.

$x=5+\sqrt{5}=7.24$ $\displaystyle x=5+\sqrt{5}=7.24$

$x=5-\sqrt{5}=2.76$ $\displaystyle x=5-\sqrt{5}=2.76$

Round to two places after the decimal.

Report an Error

Example Question #442 : Intermediate Single Variable Algebra

Solve for $\displaystyle x$ by completing the square.

$\displaystyle 2x^2-8x+1=0$

Possible Answers:

$x=2.91\text{ and }x=0.09$ $\displaystyle x=2.91\text{ and }x=0.09$

$x=8.57\text{ and }x=6.43$ $\displaystyle x=8.57\text{ and }x=6.43$

$x=1.37\text{ and }x=5.63$ $\displaystyle x=1.37\text{ and }x=5.63$

$x=3.87\text{ and }x=0.13$ $\displaystyle x=3.87\text{ and }x=0.13$

Correct answer:

$x=3.87\text{ and }x=0.13$ $\displaystyle x=3.87\text{ and }x=0.13$

Explanation:

$\displaystyle 2x^2-8x+1=0$

Start by subtracting $\displaystyle 1$ from both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle 2x^2-8x=-1$

Next, divide everything by the coefficient of the x^2 $\displaystyle x^2$ term.

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

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

Rewrite the left side of the equation in the squared form.

$(x-2)^2=\frac{7}{2}$ $\displaystyle (x-2)^2=\frac{7}{2}$

Take the square root of both sides.

$x-2=\pm \sqrt{\frac{7}{2}}$ $\displaystyle x-2=\pm \sqrt{\frac{7}{2}}$

Now solve for $\displaystyle x$.

$x=2+\sqrt{\frac{7}{2}}=3.87$ $\displaystyle x=2+\sqrt{\frac{7}{2}}=3.87$

$x=5-\sqrt{\frac{7}{2}}=0.13$ $\displaystyle x=5-\sqrt{\frac{7}{2}}=0.13$

Round to two places after the decimal.

Report an Error

Example Question #281 : Quadratic Equations And Inequalities

Solve for $\displaystyle x$ by completing the square.

$\displaystyle 3x^2-5x-10=0$

Possible Answers:

$x=3.69\text{ and }x=-2.31$ $\displaystyle x=3.69\text{ and }x=-2.31$

$x=7.08\text{ and }x=-3.92$ $\displaystyle x=7.08\text{ and }x=-3.92$

$x=4.09\text{ and }x=1.91$ $\displaystyle x=4.09\text{ and }x=1.91$

$x=2.84\text{ and }x=-1.17$ $\displaystyle x=2.84\text{ and }x=-1.17$

Correct answer:

$x=2.84\text{ and }x=-1.17$ $\displaystyle x=2.84\text{ and }x=-1.17$

Explanation:

$\displaystyle 3x^2-5x-10=0$

Start by adding $\displaystyle 10$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle 3x^2-5x=10$

Next, divide everything by the coefficient of the x^2 $\displaystyle x^2$ term.

$x^2-\frac{5}{3}x=\frac{10}{3}$ $\displaystyle x^2-\frac{5}{3}x=\frac{10}{3}$

$x^2-\frac{5}{3}x+\frac{25}{36}=\frac{10}{3}+\frac{25}{36}$ $\displaystyle x^2-\frac{5}{3}x+\frac{25}{36}=\frac{10}{3}+\frac{25}{36}$

Rewrite the left side of the equation in the squared form.

$(x-\frac{5}{6})^2=\frac{145}{36}$ $\displaystyle (x-\frac{5}{6})^2=\frac{145}{36}$

Take the square root of both sides.

$x-\frac{5}{6}=\pm \sqrt{\frac{145}{36}}$ $\displaystyle x-\frac{5}{6}=\pm \sqrt{\frac{145}{36}}$

Now solve for $\displaystyle x$.

$x=\frac{5}{6}+\sqrt{\frac{145}{36}}=2.84$ $\displaystyle x=\frac{5}{6}+\sqrt{\frac{145}{36}}=2.84$

$x=\frac{5}{6}-\sqrt{\frac{145}{36}}=-1.17$ $\displaystyle x=\frac{5}{6}-\sqrt{\frac{145}{36}}=-1.17$

Round to two places after the decimal.

Report an Error

Example Question #281 : Quadratic Equations And Inequalities

Solve for $\displaystyle x$ by completing the square.

$\displaystyle x^2-10x-12=0$

Possible Answers:

$x=12.36\text{ and }x=2.36$ $\displaystyle x=12.36\text{ and }x=2.36$

$x=-3.46\text{ and }x=-16.46$ $\displaystyle x=-3.46\text{ and }x=-16.46$

$x=20.75\text{ and }x=8.75$ $\displaystyle x=20.75\text{ and }x=8.75$

$x=11.08\text{ and }x=-1.08$ $\displaystyle x=11.08\text{ and }x=-1.08$

Correct answer:

$x=11.08\text{ and }x=-1.08$ $\displaystyle x=11.08\text{ and }x=-1.08$

Explanation:

$\displaystyle x^2-10x-12=0$

Start by adding $\displaystyle 12$ to both sides so that the terms with the $\displaystyle x$ are together on the left side of the equation.

$\displaystyle x^2-10x=12$

$\displaystyle x^2-10x+25=12+25$

Rewrite the left side of the equation in the squared form.

$\displaystyle (x-5)^2=37$

Take the square root of both sides.

$x-5=\pm \sqrt{37}$ $\displaystyle x-5=\pm \sqrt{37}$

Now solve for $\displaystyle x$.

$x=5+\sqrt{37}=11.08$ $\displaystyle x=5+\sqrt{37}=11.08$

$x=5-\sqrt{37}=-1.08$ $\displaystyle x=5-\sqrt{37}=-1.08$

Round to two places after the decimal.

Report an Error

View Algebra 2 Tutors

Lucille
Certified Tutor

View Algebra 2 Tutors

Malik
Certified Tutor

Morris College, Bachelor of Science, Mathematics. Lesley University, Masters in Education, Instructional Technology.

View Algebra 2 Tutors

Nicholas
Certified Tutor

University of St Thomas, Current Undergrad Student, Physics.

All Algebra II Resources

10 Diagnostic Tests 630 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

GMAT Tutors in New York City, English Tutors in Chicago, Spanish Tutors in Miami, Algebra Tutors in Washington DC, English Tutors in New York City, Statistics Tutors in Atlanta, Chemistry Tutors in Denver, Reading Tutors in Phoenix, LSAT Tutors in Philadelphia, SSAT Tutors in Atlanta

Popular Courses & Classes

SSAT Courses & Classes in Washington DC, SAT Courses & Classes in Phoenix, SAT Courses & Classes in Seattle, GMAT Courses & Classes in Washington DC, GRE Courses & Classes in Phoenix, ISEE Courses & Classes in Washington DC, GMAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Chicago, Spanish Courses & Classes in Seattle, SAT Courses & Classes in Miami

Popular Test Prep

SAT Test Prep in Miami, ISEE Test Prep in Houston, GMAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Chicago, GMAT Test Prep in Los Angeles, GMAT Test Prep in Chicago, GMAT Test Prep in Houston, GMAT Test Prep in Washington DC, SAT Test Prep in Los Angeles, MCAT Test Prep in Philadelphia

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

Algebra II : Algebra II

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #123 : Solving Quadratic Equations

Example Question #55 : Completing The Square

Example Question #56 : Completing The Square

Example Question #51 : Completing The Square

Example Question #1581 : Algebra Ii

Example Question #1582 : Algebra Ii

Example Question #1582 : Algebra Ii

Example Question #442 : Intermediate Single Variable Algebra

Example Question #281 : Quadratic Equations And Inequalities

Example Question #281 : Quadratic Equations And Inequalities

All Algebra II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: