ACT Math : Algebra

Study concepts, example questions & explanations for ACT Math

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Example Questions

Example Question #1 : How To Divide Complex Numbers

Simplify:

Possible Answers:

Correct answer:

Explanation:

This problem can be solved in a way similar to other kinds of division problems (with binomials, for example). We need to get the imaginary number out of the denominator, so we will multiply the denominator by its conjugate and multiply the top by it as well to preserve the number's value.

Then, recall  by definition, so we can simplify this further:

This is as far as we can simplify, so it is our final answer.

Example Question #12 : Complex Numbers

The solution of  is the set of all real numbers  such that:

 

 

Possible Answers:

Correct answer:

Explanation:

Square both sides of the equation:  

Then Solve for x:

Therefore, 

 

 

Example Question #1 : How To Multiply Complex Numbers

What is the product of  and 

Possible Answers:

Correct answer:

Explanation:

Multiplying complex numbers is like multiplying binomials, you have to use foil. The only difference is, when you multiply the two terms that have  in the them you can simplify the  to negative 1. Foil is first, outside, inside, last

First

Outside:

Inside

Last

Add them all up and you get 

Example Question #3 : How To Multiply Complex Numbers

Simplify the following:

Possible Answers:

Correct answer:

Explanation:

Begin this problem by doing a basic FOIL, treating  just like any other variable.  Thus, you know:

Recall that since .  Therefore, you can simplify further:

Example Question #3 : How To Multiply Complex Numbers

Complex numbers take the form , where  is the real term in the complex number and  is the nonreal (imaginary) term in the complex number.

Distribute: 

Possible Answers:

Correct answer:

Explanation:

This equation can be solved very similarly to a binomial like . Distribution takes place into both the real and nonreal terms inside the complex number, where applicable.

Example Question #4 : How To Multiply Complex Numbers

Complex numbers take the form , where  is the real term in the complex number and  is the nonreal (imaginary) term in the complex number.

Distribute and solve: 

Possible Answers:

Correct answer:

Explanation:

This problem can be solved very similarly to a binomial like .

Example Question #22 : Complex Numbers

Complex numbers take the form , where  is the real term in the complex number and  is the nonreal (imaginary) term in the complex number.

Which of the following is equivalent to ?

Possible Answers:

Correct answer:

Explanation:

When dealing with complex numbers, remember that .

If we square , we thus get .

Yet another exponent gives us  OR .

But when we hit , we discover that 

Thus, we have a repeating pattern with powers of , with every 4 exponents repeating the pattern. This means any power of  evenly divisible by 4 will equal 1, any power of  divisible by 4 with a remainder of 1 will equal , and so on.

 

Thus, 

 

Since the remainder is 3, we know that .

Example Question #3 : How To Multiply Complex Numbers

Simplify the following:

Possible Answers:

Correct answer:

Explanation:

Begin by treating this just like any normal case of FOIL. Notice that this is really the form of a difference of squares. Therefore, the distribution is very simple. Thus:

Now, recall that . Therefore,  is . Based on this, we can simplify further:

Example Question #8 : How To Multiply Complex Numbers

Which of the following is equal to ?

Possible Answers:

Correct answer:

Explanation:

Remember that since , you know that  is .  Therefore,  is  or .  This makes our question very easy.

 is the same as  or 

Thus, we know that  is the same as  or .

Example Question #41 : Squaring / Square Roots / Radicals

Complex numbers take the form , where  is the real term in the complex number and  is the nonreal (imaginary) term in the complex number.

Simplify the following expression, leaving no complex numbers in the denominator.

Possible Answers:

Correct answer:

Explanation:

Solving this problem requires eliminating the nonreal term of the denominator. Our best bet for this is to cancel the nonreal term out by using the conjugate of the denominator.

Remember that for all binomials , there exists a conjugate  such that .

This can also be applied to complex conjugates, which will eliminate the nonreal portion entirely (since )!

  Multiply both terms by the denominator's conjugate.

 Simplify. Note .

 FOIL the numerator.

 Combine and simplify.

 Simplify the fraction.

Thus, .

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