We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

SAT Math : Algebra

Study concepts, example questions & explanations for SAT Math

Create An Account

All SAT Math Resources

16 Diagnostic Tests 660 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : How To Divide Complex Numbers

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(2+i) \ast i$ .

Possible Answers:

$2+ \frac{ 6 }{5}i$

$\frac{ 2 }{3}+2i$

$\frac{ 6 }{5} + \frac{ 2 }{5}i$

$\frac{ 2 }{5} + \frac{ 6 }{5}i$

$2+ \frac{ 2 }{3}i$

Correct answer:

$\frac{ 6 }{5} + \frac{ 2 }{5}i$

Explanation:

$a \ast b = \frac{a + b}{a}$ , so

$(2+i) \ast i = \frac{(2+ i )+ i}{2+i} = \frac{2 + 2i}{2+i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is 2 - i :

$(2+i) \ast i$

$= \frac{2 + 2i}{2+i}$

$= \frac{\left (2 + 2i \right )\left (2-i \right )}{\left (2+i \right )\left (2-i \right )}$

$= \frac{ 2 \cdot 2 - 2 \cdot i + 2i \cdot 2 - 2i \cdot i}{2 ^{2}+1^{2}}$

$= \frac{ 4 - 2 i + 4i - 2 \cdot i^{2}}{4+1}$

$= \frac{ 4 - 2 i + 4i - 2 \cdot (-1)}{5}$

$= \frac{ 4 - 2 i + 4i +2}{5}$

$= \frac{ 6 + 2 i }{5}$

$= \frac{ 6 }{5} + \frac{ 2 }{5}i$

Report an Error

Example Question #602 : Algebra

Define an operation $\Phi$ such that, for any complex number ,

$\Phi a = a + 2ai$

If $\Phi N = i$ , then evaluate .

Possible Answers:

$\frac{ 2}{5} + \frac{1}{5} i$

$\frac{ 1}{3} + \frac{2}{3} i$

$\frac{ 2}{5} - \frac{1}{5} i$

$\frac{ 2}{3} + \frac{1}{3} i$

$\frac{ 1}{5} - \frac{2}{5} i$

Correct answer:

$\frac{ 2}{5} + \frac{1}{5} i$

Explanation:

$\Phi a = a + 2ai$ , so

$\Phi N = N + 2Ni = N (1+ 2i)$

$\Phi N = i$ , so

N (1+ 2i) = i , and

$N=\frac{ i}{ 1+ 2i}$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is 1 -2i :

$N=\frac{ i}{ 1+ 2i}$

$=\frac{ i( 1- 2i)}{( 1+ 2i)( 1- 2i)}$

$=\frac{ i \cdot 1- i \cdot 2i}{1 ^{2}+2^{2}}$

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

$=\frac{ i - 2(-1)}{5}$

$=\frac{ i +2}{5}$

$=\frac{ 2}{5} + \frac{1}{5} i$

Report an Error

Example Question #3 : How To Divide Complex Numbers

Define an operation $\blacklozenge$ as follows:

For any two complex numbers and ,

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$

Evaluate $(1+ i) \blacklozenge (1+ i)$ .

Possible Answers:

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

$\frac{ 1 }{2} + \frac{3}{2} i$

$\frac{ 3 }{2} + \frac{1}{2} i$

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

Correct answer:

$\frac{ 3 }{2} + \frac{1}{2} i$

Explanation:

$a \blacklozenge b = \frac{a}{2i} + \frac{2i}{b}$ , so

$(1+ i) \blacklozenge (1+ i) = \frac{1+ i}{2i} + \frac{2i}{1+ i}$

We can simplify each expression separately by rationalizing the denominators.

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

$= \frac{\left (1+ i \right )\cdot i}{2i \cdot i}$

$= \frac{1 \cdot i+ i \cdot i}{2 \cdot i ^{2}}$

$= \frac{ i+(-1)}{2 (-1)}$

$= \frac{-1+i}{-2}$

$= \frac{ 1 }{2} - \frac{1}{2} i$

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

$= \frac{2i(1- i)}{(1+ i)(1- i)}$

$= \frac{2i \cdot 1- 2i \cdot i}{1^{2}+1^{2}}$

$= \frac{2i - 2i ^{2}}{1+1}$

$= \frac{2i - 2(-1)}{2}$

$= \frac{2i +2}{2}$

= 1 + i

Therefore,

$(1+ i) \blacklozenge (1+ i) = \frac{1+ i}{2i} + \frac{2i}{1+ i}$

$=\left ( \frac{ 1 }{2} - \frac{1}{2} i \right )+ \left (1 + i \right )$

$=\left ( \frac{ 1 }{2}+1 \right )+ \left ( i- \frac{1}{2} i \right )$

$= \frac{ 3 }{2} + \frac{1}{2} i$

Report an Error

Example Question #11 : Complex Numbers

Define an operation $\ast$ so that for any two complex numbers and :

$a \ast b = \frac{a + b}{a}$

Evaluate $(3+2i) \ast 2$

Possible Answers:

$\frac{11 }{13} + \frac{ 16}{13}i$

$\frac{19 }{13} - \frac{ 4}{13}i$

$\frac{19}{13} + \frac{ 16}{13}i$

$\frac{11 }{13} - \frac{ 4}{13}i$

$\frac{19}{13} - \frac{ 16}{13}i$

Correct answer:

$\frac{19 }{13} - \frac{ 4}{13}i$

Explanation:

$a \ast b = \frac{a + b}{a}$ , so

$(3+2i) \ast 2 = \frac{(3+2i) + 2}{3+2i } = \frac{5+2i}{3+2i }$

Rationalize the denominator by multiplying both numerator and denominator by the complex conjugate of the latter, which is 3 -2i :

$(3+2i) \ast 2 = \frac{5+2i}{3+2i }$

$= \frac{\left (5+2i \right )\left (3-2i \right )}{\left (3+2i \right )\left (3-2i \right )}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2i \cdot 2i }{3^{2}+2^{2}}$

$= \frac{5 \cdot 3 - 5 \cdot 2i + 2i \cdot 3 - 2 \cdot 2 \cdot i^{2} }{9+4}$

$= \frac{15 - 10i + 6i -4 \cdot (-1) }{9+4}$

$= \frac{15 - 10i + 6i +4}{13}$

$= \frac{19 -4i}{13}$

$= \frac{19 }{13} - \frac{ 4}{13}i$

Report an Error

Example Question #2381 : Sat Mathematics

Define an operation $\Theta$ such that for any complex number ,

$\Theta a = 2a i + i$

If $\Theta N = 1,000$ , evaluate .

Possible Answers:

$500 + \frac{1}{2}i$

$\frac{ 1 }{2}+ 500i$

$- \frac{ 1 }{2}- 500i$

$- \frac{ 1 }{2}+ 500i$

$500 - \frac{1}{2}i$

Correct answer:

$- \frac{ 1 }{2}- 500i$

Explanation:

First substitute our variable N in where ever there is an a.

Thus, $\Theta a = 2a i + i$ , becomes $\Theta N = 2N i + i$ .

Since $\Theta N = 1,000$ , substitute:

2N i + i = 1,000

In order to solve for the variable we will need to isolate the variable on one side with all other constants on the other side. To do this, apply the oppisite operation to the function.

First subtract i from both sides.

2N i= 1,000 - i

Now divide by 2i on both sides.

$\frac{2N i}{2i}= \frac{1,000 - i}{2i}$

From here multiply the numerator and denominator by i to further solve.

$N= \frac{1,000 - i}{2i}$

$= \frac{(1,000 - i) \cdot i }{2i \cdot i}$

$= \frac{1,000\cdot i - i\cdot i }{2 \cdot i^{2}}$

Recall that i^2=-1 by definition. Therefore,

$= \frac{1,000 i - (-1) }{2 \cdot (-1)}$

$= \frac{1,000 i +1 }{-2}$

$= \frac{ 1 }{-2}+ \frac{1,000 i }{-2}$

$=- \frac{ 1 }{2}- 500i$ .

Report an Error

Example Question #601 : Algebra

Let 4x^2 = y^3 . What is the following equivalent to, in terms of :

$\frac{x}{\sqrt{y}}$

Possible Answers:

$\pm\frac{1}{8}y^{\frac{3}{2}}$

$\frac{1}{2}\sqrt{y}$

$\pm\frac{1}{2}y$

$\frac{1}{4}y^2$

Correct answer:

$\pm\frac{1}{2}y$

Explanation:

Solve for x first in terms of y, and plug back into the equation.

$\\4x^2 = y^3 \\ \\x^2 = \frac{1}{4} y^3 \\ \\x = \pm \frac{1}{2} y\sqrt{y}$

Then go back to the equation you are solving for:

$\frac{x}{\sqrt{y}}$ substitute in $x=\pm\frac{1}{2}y\sqrt{y}$

$\\ \frac{x}{\sqrt{y}}=\frac{\pm\frac{1}{2}y\sqrt{y}}{\sqrt{y}} \\ \\\frac{x}{\sqrt{y}}=\pm\frac{1}{2}y$

Report an Error

Example Question #602 : Algebra

Simplify the expression by rationalizing the denominator, and write the result in standard form: $\frac{5+i}{7-i}$

Possible Answers:

$\frac{18}{25} + \frac{ 6}{25}i$

$\frac{3}{4} + \frac{ 1}{4}i$

$\frac{17}{25} + \frac{ 6}{25}i$

$\frac{17}{24} + \frac{ 1}{4}i$

$\left (\textup{Note: } i=\sqrt{-1}\right\ )$

Correct answer:

$\frac{17}{25} + \frac{ 6}{25}i$

Explanation:

Multiply both numerator and denominator by the complex conjugate of the denominator, which is 7 + i :

$\frac{5+i}{7-i}$

$= \frac{(5+i)(7+i)}{(7-i)(7+i)}$

$= \frac{5 \cdot 7 + 5 \cdot i + 7 \cdot i + i \cdot i }{(7^{2}+1 ^{2}) }$

$= \frac{35 +5i + 7 i +(-1) }{(49+1) }$

$= \frac{34 +12i }{50 }$

$= \frac{34 }{50 } + \frac{ 12i }{50 }$

$= \frac{17}{25} + \frac{ 6}{25}i$

Report an Error

Example Question #603 : Algebra

Find the product of (3 + 4i)(4 - 3i) given that i is the square root of negative one.

Possible Answers:

7 + i

24 + 7i

12 - 12i

Correct answer: 24 + 7i

Explanation:

Distribute (3 + 4i)(4 - 3i)

3(4) + 3(-3i) + 4i(4) + 4i(-3i)

12 - 9i + 16i -12i²

12 + 7i - 12(-1)

12 + 7i + 12

24 + 7i

Report an Error

Example Question #604 : Algebra

$x^{4}-1$ has 4 roots, including the complex numbers. Take the product of $\left ( 5+2i \right )$ with each of these roots. Take the sum of these 4 results. Which of the following is equal to this sum?

Possible Answers:

3+7i

4-10i

The correct answer is not listed.

10i

Correct answer:

Explanation:

$x^{4}-1=\left ( x^{2} -1\right )\left ( x^{2}+1 \right )=\left ( x+1 \right )\left ( x-1 \right )\left ( x-i \right )\left ( x+i \right )$

This gives us roots of

$1,\ -1,\ i,\ -i$

The product of $\left ( 5+2i \right )$ with each of these gives us:

$(5+2i)\cdot i = 5i + 2i^2 = 5i +2(-1) = 5i-2 = -2 + 5i$

$(5+2i)\cdot (-i) = -5i -2i^2 = -5i -2(-1) = -5i +2 = 2-5i$

$(5+2i)\cdot 1 = 5+2i$

$(5+2i)\cdot (-1) = -5 -2i$

The sum of these 4 is:

[(-2+5i) + (2-5i)] + [(5+2i) + (-5-2i)] =

=[(-2)+2 +5i - 5i] + [5-5 + 2i - 2i] = [0] + [0] = 0

What we notice is that each of the roots has a negative. It thus makes sense that they will all cancel out. Rather than going through all the multiplication, we can instead look at the very beginning setup, which we can simplify using the distributive property:

$1(5+2i) + (-1)\cdot(5+2i)+ i(5+2i)+(-i) \cdot(5+2i) = (5+2i)\cdot[1+(-1)+i+(-i)]$

$= (5+2i)\cdot[0] = 0$

Report an Error

Example Question #2386 : Sat Mathematics

Simplify:

$(3i)^{6}$

Possible Answers:

729 i

-729 i

None of the other responses gives the correct answer.

729

Correct answer:

Explanation:

Apply the Power of a Product Property:

$(3i)^{6} = 3 ^{6} \cdot i^{6} = 729 \cdot i^{6}$

A power of can be found by dividing the exponent by 4 and noting the remainder. 6 divided by 4 is equal to 1, with remainder 2, so

$i^{6} = i^{2} = -1$

Substituting,

$(3i)^{6} = 729 \cdot (-1) = -729$ .

Report an Error

View SAT Mathematics Tutors

Amanda
Certified Tutor

Seattle University, Current Undergrad Student, Applied Mathematics.

View SAT Mathematics Tutors

Sharif
Certified Tutor

NCCU, Bachelors, Physics and Mathematics.

View SAT Mathematics Tutors

Joy
Certified Tutor

Boston University, Bachelor in Arts, Conservation Biology.

All SAT Math Resources

16 Diagnostic Tests 660 Practice Tests Question of the Day Flashcards Learn by Concept

SAT Math Tutors in Top Cities:

Atlanta SAT Math Tutors, Austin SAT Math Tutors, Boston SAT Math Tutors, Chicago SAT Math Tutors, Dallas Fort Worth SAT Math Tutors, Denver SAT Math Tutors, Houston SAT Math Tutors, Kansas City SAT Math Tutors, Los Angeles SAT Math Tutors, Miami SAT Math Tutors, New York City SAT Math Tutors, Philadelphia SAT Math Tutors, Phoenix SAT Math Tutors, San Diego SAT Math Tutors, San Francisco-Bay Area SAT Math Tutors, Seattle SAT Math Tutors, St. Louis SAT Math Tutors, Tucson SAT Math Tutors, Washington DC SAT Math Tutors

Popular Courses & Classes

LSAT Courses & Classes in Chicago, GRE Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Philadelphia, GRE Courses & Classes in Atlanta, Spanish Courses & Classes in New York City, GMAT Courses & Classes in Seattle, ACT Courses & Classes in Chicago, MCAT Courses & Classes in Denver, MCAT Courses & Classes in Los Angeles, GMAT Courses & Classes in Houston

Popular Test Prep

SSAT Test Prep in New York City, LSAT Test Prep in New York City, MCAT Test Prep in Phoenix, GMAT Test Prep in Los Angeles, MCAT Test Prep in Chicago, LSAT Test Prep in Chicago, GMAT Test Prep in Houston, SAT Test Prep in San Diego, GRE Test Prep in Phoenix, SSAT Test Prep in Houston

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

SAT Math : Algebra

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #2 : How To Divide Complex Numbers

Example Question #602 : Algebra

Example Question #3 : How To Divide Complex Numbers

Example Question #11 : Complex Numbers

Example Question #2381 : Sat Mathematics

Example Question #601 : Algebra

Example Question #602 : Algebra

Example Question #603 : Algebra

Example Question #604 : Algebra

Example Question #2386 : Sat Mathematics

All SAT Math Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: