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 #144 : Understanding Quadratic Equations

Let

(x-3)^2 = 36

Determine the value of x.

Possible Answers:

x=6,-6

x=6

x=-3, 9

x=9

Correct answer:

x=-3, 9

Explanation:

To solve for x we need to isolate x. We can do this by taking the square root of each side and then doing algebraic operations.

(x-3)^2=36

$\sqrt(x-3)^2=\sqrt36$

$x-3=\pm6$

Now we need to separate our equation in two and solve for each x.

x-3=6 or x-3=-6

x=9 x=-3

Report an Error

Example Question #21 : Linear / Rational / Variable Equations

Solve for :

$(x-3)^{2}=36$

Possible Answers:

(3,-9)

None of the other answers

(6,-6)

(9,-3)

(9,3)

Correct answer:

(9,-3)

Explanation:

To solve this equation, you must first eliminate the exponent from the (x-3) by taking the square root of both sides:

$\sqrt{(x-3)^{2}}=\sqrt{36}$

Since the square root of 36 could be either or , there must be 2 values of . So, solve for

x-3=-6

and

x-3=6

to get solutions of (9,-3) .

Report an Error

Example Question #1 : Quadratic Roots

Find the roots of $y = x^{2} + 9x + 14$ .

Possible Answers:

x = 4, x = 5

x=0

x = 7, x = 2

x = -7, x = -2

x = 14, x = 1

Correct answer:

x = -7, x = -2

Explanation:

When we factor, we are looking for two number that multiply to the constant, , and add to the middle term, . Looking through the factors of , we can find those factors to be and .

Thus, we have the factors:

(x + 2)(x+7) .

To solve for the solutions, set each of these factors equal to zero.

Thus, we get x + 2 = 0 , or x = -2 .

Our second solution is, x + 7 =0 , or x = -7 .

Report an Error

Example Question #311 : Intermediate Single Variable Algebra

Write a quadratic function in standard form with roots of -1 and 2.

Possible Answers:

f(x)=x^2+x-2

f(x)=x^2-2x-2

f(x)=x^2-3x-2

f(x)=x^2-x-2

f(x)=x^2-x+2

Correct answer:

f(x)=x^2-x-2

Explanation:

From the zeroes we know

f(x) = (x+1)(x-2)

Use FOIL method to obtain:

f(x)=x^2-x-2

Report an Error

Example Question #1453 : Algebra Ii

Select the quadratic equation that has these roots:

(2x+5)(x-2)

Possible Answers:

None of these.

2x^2+x-10

2x^2-x+10

2x^2-x-10

2x^2+x+10

Correct answer:

2x^2+x-10

Explanation:

(2x+5)(x-2)

FOIL the two factors to find the quadratic equation.

First terms:

$2x\cdot x=2x^2$

Outer terms:

$2x\cdot -2=-4x$

Inner terms:

$5\cdot x=5x$

Last terms:

$5\cdot -2=-10$

Simplify:

2x^2-4x+5x-10

$\boldsymbol{2x^2+x-10}$

Report an Error

Example Question #1454 : Algebra Ii

Solve for a possible root: y=3x^2-6x+8

Possible Answers:

$\textup{There are no possible roots.}$

6+6i

$1-\frac{4i\sqrt{3}}{3}$

$1-\frac{\sqrt{25}}{2}$

$1-\frac{i\sqrt{15}}{3}$

Correct answer:

$1-\frac{i\sqrt{15}}{3}$

Explanation:

Write the quadratic equation.

$x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

The equation y=3x^2-6x+8 is in the form y=ax^2+bx+c .

Substitute the proper coefficients into the quadratic equation.

$x= \frac{-(-6)\pm\sqrt{(-6)^2-4(3)(8)}}{2(3)}=\frac{6\pm\sqrt{36-96}}{6}$

$x=\frac{6\pm\sqrt{36-96}}{6} =\frac{6\pm\sqrt{-60}}{6}$

The negative square root can be replaced by the imaginary term . Simplify square root 60 by common factors of numbers with perfect squares.

$\frac{6\pm\sqrt{-60}}{6} = \frac{6\pm i\sqrt{60}}{6} =\frac{6\pm i\sqrt{4} \cdot \sqrt{15}}{6} =\frac{6\pm 2i\sqrt{15}}{6}$

Simplify the fraction.

$\frac{6\pm 2i\sqrt{15}}{6} = 1\pm \frac{i\sqrt{15}}{3}$

A possible root is: $1-\frac{i\sqrt{15}}{3}$

Report an Error

Example Question #1455 : Algebra Ii

Solve for the roots (if any) of y=-8x^2+28x-12

Possible Answers:

$-\frac{1}{2},3$

$\textup{There are no real roots.}$

$\frac{1}{2},3$

$\frac{1}{2},\frac{5}{2}$

$-\frac{1}{4},5$

Correct answer:

$\frac{1}{2},3$

Explanation:

Pull out a common factor of negative four.

y=-4(2x^2-7x+3)

The term inside the parentheses can be factored.

2x^2-7x+3 = (2x-1)(x-3)

Set the binomials equal to zero and solve for the roots. We can ignore the negative four coefficient.

$2x-1=0, x=\frac{1}{2}$

x-3=0, x=3

The answers are: $\frac{1}{2},3$

Report an Error

Example Question #1 : Finding Roots

$y=x^{2}-2x-3$

Factor the above function to find the roots of the quadratic equation.

Possible Answers:

x=-3,1

x=-3,-2

x=-1,3

x=2,3

Correct answer:

x=-1,3

Explanation:

Factoring a quadratic equation means doing FOIL backwards. Recall that when you use FOIL, you start with two binomials and end with a trinomial:

$(x+a)(x+b)=x^{2}+(a+b)x+ab$

Now, we're trying to go the other direction -- starting with a trinomial, and going back to two factors.

Here, -3 is equal to , and -2 is equal to a+b . We can use this information to find out what and are, separately. In other words, we have to find two factors of -3 that add up to -2.

Factors of -3:

3*-1 (sum = 2)
-3*1 (sum = -2)

Thus our factored equation should look like this:

y=(x-3)(x+1)

The roots of the quadratic equation are the values of x for which y is 0.

0=(x-3)(x+1)

We know that anything times zero is zero. So the entire expression equals zero when at least one of the factors equals zero.

$x-3=0\: \: \textup{and} \: \: x+1=0$

$x=3 \: \: \textup{and}\: \: -1$

Report an Error

Example Question #1 : Finding Zeros Of A Polynomial

Find the roots of the function:
$\small f(x)=x^2+4x-12$

Possible Answers:

$\small x=-4\ or\ 3$

$\small \small x=2\ or\ -6$

$\small x=-3\ or\ 4$

$\small \small x=-2\ or\ 6$

Correct answer:

$\small \small x=2\ or\ -6$

Explanation:

Factor:

0=x^2+4x-12

$\small \small 0=(x-2)(x+6)$

Double check by factoring:

$\small F: (x)(x)$

$\small \small O:(x)(6)=6x$

$\small \small I: (-2)(x)=-2x$

$\small \small L: (-2)(6)=-12$

Add together: $\small x^2+6x-2x-12=x^2+4x-12$

Therefore:

$\small x-2=0\ or x+6=0$

$\small x=2\or-6$

Report an Error

Example Question #2 : Finding Roots

Solve for x.

$x^{2}-7x+10=0$

Possible Answers:

x = 5, 2

x = –5, –2

x = 4, 3

x = –4, –3

x = 5

Correct answer:

x = 5, 2

Explanation:

1) Split up the middle term so that factoring by grouping is possible.

Factors of 10 include:

1 * 10= 10 1 + 10 = 11

2 * 5 =10 2 + 5 = 7

–2 * –5 = 10 –2 + –5 = –7 Good!

$x^{2}-2x-5x+10=0$

2) Now factor by grouping, pulling "x" out of the first pair and "-5" out of the second.

x(x-2)-5(x-2)=0

3) Now pull out the common factor, the "(x-2)," from both terms.

(x-5)(x-2)=0

4) Set both terms equal to zero to find the possible roots and solve using inverse operations.

x – 5 = 0, x = 5

x – 2 = 0, x = 2

Report an Error

View Algebra 2 Tutors

Melody
Certified Tutor

George Fox University, Bachelor of Science, Biology, General.

View Algebra 2 Tutors

Kevin
Certified Tutor

St. Johns University, Bachelors, Math/Coaching.

View Algebra 2 Tutors

Peter
Certified Tutor

Nottingham University UK, Bachelor in Arts, Mathematics.

All Algebra II Resources

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

Popular Subjects

ACT Tutors in Phoenix, Computer Science Tutors in Washington DC, SAT Tutors in Chicago, GRE Tutors in Dallas Fort Worth, Math Tutors in Washington DC, Biology Tutors in Boston, Calculus Tutors in Houston, Calculus Tutors in Phoenix, English Tutors in Denver, LSAT Tutors in Dallas Fort Worth

Popular Courses & Classes

MCAT Courses & Classes in Phoenix, MCAT Courses & Classes in Philadelphia, Spanish Courses & Classes in Dallas Fort Worth, ACT Courses & Classes in New York City, GMAT Courses & Classes in Boston, LSAT Courses & Classes in Denver, Spanish Courses & Classes in Houston, LSAT Courses & Classes in Philadelphia, ACT Courses & Classes in Denver, MCAT Courses & Classes in Boston

Popular Test Prep

ISEE Test Prep in Phoenix, LSAT Test Prep in Denver, ACT Test Prep in Miami, SAT Test Prep in Boston, SSAT Test Prep in New York City, LSAT Test Prep in New York City, MCAT Test Prep in San Diego, ISEE Test Prep in Boston, MCAT Test Prep in Houston, GMAT Test Prep in Dallas Fort Worth

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 #144 : Understanding Quadratic Equations

Example Question #21 : Linear / Rational / Variable Equations

Example Question #1 : Quadratic Roots

Example Question #311 : Intermediate Single Variable Algebra

Example Question #1453 : Algebra Ii

Example Question #1454 : Algebra Ii

Example Question #1455 : Algebra Ii

Example Question #1 : Finding Roots

Example Question #1 : Finding Zeros Of A Polynomial

Example Question #2 : Finding Roots

All Algebra II Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: