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Chris Lele

SAT Passport to Advanced Math, Part I

The SAT has provided the whimsical (and slightly intimidating) title “Passport to Advanced Math”. The good news is that it is actually not advanced math. It’s actually the skills that underpin some of the more advanced stuff you’ll be doing in calculus and pre-calculus. In this light the name passport doesn’t seem as fanciful, since in order to travel in the land of advanced math you’ll need a “passport” showing that you have the fundamentals down.

So what exactly are these skills required to approach advanced math? Well, the New SAT wants you to be able to deal not just with linear equations, as you do in the Heart of Algebra section, but polynomials. However, we are talking about not your basic quadratic, no power higher than 2, but polynomials in which the powers can be very high, as in the following:

5x^4+6x^3- 2cx^2+x+1

That doesn’t mean quadratics will be absent, however most will not be your garden variety x^2-2x+1 kind and will require you to use the quadratic formula (yes, you’ll have to memorize that).

Other topics include the graphs of complex polynomials. The point of the test, though, isn’t to throw ridiculously complex equations at you; it wants to test your ability to recognize simple patterns in all that complexity. In other words, can you navigate around all the noise to find a relatively straightforward approach to solving the question?

Passport to Advanced Math Practice Question 1

Which of the following is equivalent to 9a^8 - 4a^4?
A. a^4 (9a^4-4a)
B. (3a^4-2a^4)(3a^4+2a^4)
C. (3a^4-2a^2)(3a^4+2a^2)
D. a^4 (3a-2)(3a+2)


The key here is noticing that the form (x - y)(x + y), or x^2 - y^2 can be applied to the original equation, since both ‘9’ and ‘4’ are perfect squares and the exponents are even integers, giving us (C).

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Passport to Advanced Math Practice Question 2

x+4=sqrt{(x+6 )}
What is the solution set for the above equation?
A) {-2}
B) {-2, -5}
C) {-5}
D) No solutions.
To solve for ‘x’ we want to remove the square root sign. To do so, we have to square both sides, giving us:


Balancing the equation gives us:


x = -5 and -2

This seems like a pretty straightforward question, but here’s the twist: whenever you have a square root sign over the variable one side of the equation, watch out for the quantity under the variable equaling a negative when you plug the value back in. The reason is if you get a negative number underneath the square root sign, you do not have a valid solution.

Another possibility in which one of the ‘x’s you solve for is not valid, is if one side of the equation is a square root sign and the other side of the equation (the one that does not have a square root sign) is a negative number. The reason is that the square root of a negative number will never equal a negative number.

This is what is going on here, since when you plug in x = -5 into the original equation you end up getting the following:

-1 = sqrt{1}

This is not valid, so therefore -5 is not an answer. Only -2 is, giving us A).

Passport to Advanced Math Practice Question 3

Student-Produced Response:

If g(x) = x + 5 and f(x) = 2g(x) - 1, what is f(3)?
Functions, or the application of a certain “rule”, falls under Passport to Advanced math. These question types can take on many different varieties. I’d recommend practicing these if you are not comfortable with them.

For this question, when you see f(3) that translates loosely to the following: where you see ‘x’ on the right side of the equals sign, place a ‘3’. In this case, the ‘x’ is inside the ‘g’. So you end up getting:

f(3) = 2g(3) - 1

Since g(x) = x + 5, where you see ‘x’, place a 3 forall g(3) = 3 + 5 = 8.

Knowing that g(3) = 8, we go back to the original equation:

f(x) = 2g(3) - 1 =

f(x) = 2(8) - 1 = 15, which is the answer.


About Chris Lele

Chris Lele is the GRE and SAT Curriculum Manager (and vocabulary wizard) at Magoosh Online Test Prep. In his time at Magoosh, he has inspired countless students across the globe, turning what is otherwise a daunting experience into an opportunity for learning, growth, and fun. Some of his students have even gone on to get near perfect scores. Chris is also very popular on the internet. His GRE channel on YouTube has over 10 million views. You can read Chris's awesome blog posts on the Magoosh GRE blog and High School blog! You can follow him on Twitter and Facebook!

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