All SAT Mathematics Resources
Example Questions
Example Question #2 : Exponents & Roots
If the expression is rewritten in the form , where and are real numbers, what is the value of ? (Note: )
The twist in this question is that you must be sure to answer the question they are actually asking you! The question does not ask you to provide the entire value of the complex number that you get from solving the division problem--it only asks for the real number part of that complex number, which is called . (Please note: Even though itself is also a real number, as the question states, in the expression this value is the coefficient of , so it is part of the imaginary number part of the expression.) Therefore, after you perform the division of complex numbers correctly, you have to focus only on the real number part of your answer, not on the imaginary number part.
But first, you do have to perform the division of complex numbers correctly. The most difficult and critical key step of this process is the very first step: You must multiply both the numerator and denominator by what we call the complex conjugate of the denominator, which is a fancy term that just means you change the sign in the middle of the complex number (before the imaginary part). The reason we do this is that this will make the imaginary part in the denominator disappear! Recall the difference of squares formula in algebra: . This formula works because when you FOIL , the two middle terms will cancel each other out. Well, the same thing happens here when you multiply the complex number in the denominator by its complex conjugate: the imaginary number middle terms will cancel each other out! So . The middle terms cancel each other out, leaving us with . Now you have to know that because , therefore . So , and now the negative signs in the second term cancel out, making it positive: . The point of this whole process is that the end result is to simplify the denominator to a single real number, which in this case is .
To continue the complex number division process, now you must also multiply the original numerator by the complex conjugate of the denominator: . The two middle terms, the imaginary parts, combine: . Continuing, you must know that as we stated above, so . Finishing the simplification of the numerator, we get .
Conveniently, we now see that our numerator and our denominator (see above) simplify dramatically! . Wow! (Note: On the SAT, these things never happen by accident. The test question writers carefully design the questions so that they often simplify dramatically like this, when the student solves the questions correctly.)
Now, by solving the complex number division process correctly, we have rewritten the original expression in the form , which in this case has the value . Recall that you must answer the question they are actually asking you, and the question asks you for the value of . Therefore the correct answer choice is .
Pro Tip: Experienced math students may be able to save a little time during the solution process: If you know that the question is only asking for the real number part of the final complex number, then when you solve the numerator above, you can skip the imaginary number parts, which are the two middle terms when you FOIL . So you can just solve the terms that will become the real number parts: . Since this is the real number part of the numerator, and the whole denominator is also , now you already know the final real number part will be , which is what the question is actually asking for, so that is the final correct answer choice for this question.
Example Question #7 : Working With Imaginary Numbers
If the expression is rewritten in the form , where and are real numbers, what is the value of ? (Note: )
The twist in this question is that you must be sure to answer the question they are actually asking you! The question does not ask you to provide the entire value of the complex number that you get from solving the division problem--it only asks for the coefficient of the imaginary number part of that complex number, which is called . (Please note: Even though itself is a real number, as the question states, in the expression this value b is the coefficient of , so it is part of the imaginary number part of the expression.) Therefore, after you perform the division of complex numbers correctly, you have to focus only on the coefficient of the imaginary number part of your answer, not on the real number part.
But first you do have to perform the division of complex numbers correctly. The most difficult and critical key step of this process is the very first step: You must multiply both the numerator and denominator by what we call the complex conjugate of the denominator, which is a fancy term that just means you change the sign in the middle of the complex number (before the imaginary part). The reason we do this is that this will make the imaginary part in the denominator disappear! Recall the difference of squares formula in algebra: . This formula works because when you FOIL , the two middle terms will cancel each other out. Well, the same thing happens here when you multiply the complex number in the denominator by its complex conjugate: the imaginary number middle terms will cancel each other out! So . The middle terms cancel each other out, leaving us with . Now you have to know that because , therefore . So , and now the negative signs in the second term cancel out, making it positive: . The point of this whole process is that the end result is to simplify the denominator to a single real number, which in this case is . To continue the complex number division process, now you must also multiply the original numerator by the complex conjugate of the denominator: . The two middle terms, the imaginary parts, combine: . Continuing, you must know that as we stated above, so . Finishing the simplification of the numerator, we get .
Conveniently, we now see that our numerator and our denominator (see above) simplify dramatically! . Wow! (Note: On the SAT, these things never happen by accident. The test question writers carefully design the questions so that they often simplify dramatically like this, when the student solves the questions correctly.) Now, by solving the complex number division process correctly, we have rewritten the original expression in the form which in this case has the value . Recall that you must answer the question they are actually asking you, and the question asks you for the value of . Therefore the correct answer choice is . (Note: “” is the same value as “”, so the coefficient is , even though the number is usually not written before the “” in this case since it is not necessary.)
Pro Tip: Experienced math students may be able to save a little time during the solution process: If you know that the question is only asking for the coefficient of the imaginary number part of the final complex number, then when you solve the numerator above, you can skip the real number parts, which are the first and last terms when you FOIL . So you can just solve the terms that will become the imaginary number parts: . Since this is the imaginary number part of the numerator, and the whole denominator is , now you already know the final imaginary number part will be , and its coefficient is , which is what the question is actually asking for, so that is the final correct answer choice for this question.
Example Question #1 : Solving Problems With Roots
Simplify
For what value of is this equation true?
To solve for , we must first square both sides to get rid of the radical. We get
We subtract both sides by to get the alone.
We square root both sides to get
Answer choice and are incorrect.
Answer choice is incorrect because it was not square rooted.
Example Question #2 : Solving Problems With Roots
Simplify
For what value of is this equation true?
To solve for , we must first square both sides to get rid of the radical. We get . We subtract both sides by to get the alone.
We divide by to get alone .
We square root both sides to get Since is not listed as an answer choice, we simplify. The highest square root that can multiply to is . We take the out of the radical to get .
Example Question #3 : Solving Problems With Roots
Simplify:
To solve this problem we must first simplify the radical by breaking it up into two parts becomes then we simplify into to get
We multiply to get , then divide by to get
Example Question #4 : Solving Problems With Roots
Find the value of
To solve this problem we must first simplify into and further into
Then we can multiply to get
To find we first cancel out the on both sides and then divide by and get
Example Question #5 : Solving Problems With Roots
Find the value of
To solve this problem we must first subtract from both sides
Then we square both sides
Add the to both sides
Divide both sides by
Example Question #6 : Solving Problems With Roots
Find the value of
To solve this problem we first multiply both sides by to get rid of the fraction
Then we add to both sides
We move to the left side to set the equation equal to . This way we are able to factor the equation as if it was a quadratic.
And now we can factor into
Therefore the value of is
does not exist
Example Question #7 : Solving Problems With Roots
Find the value of
To solve this problem we first multiply both sides by to get rid of the fraction
Then we add to both sides
We move to the left side to set the equation equal to . This way we are able to factor the equation as if it was a quadratic.
And now we can factor into
Therefore the value of is
, does not exist
,
Example Question #8 : Solving Problems With Roots
Find the value of
Only
and
and
Only
and
To solve this problem we must first subtract a square from both sides
We move to the right side to set the equation equal to . This way we are able to factor the equation as if it was a quadratic.
And now we can factor into
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