Solving 10 Of The Hardest SAT Math Problems from Official Tests

[@preppros]

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[@RickEs_Pi]

I did the last one differently without knowing the rule (x+y)(x-y) = x^2-y^2. I know that x^4 + y^4 = 225 (by exponentiation to 2 in x^2 +y^2 = 15). I also know that the result of x^2-y^2 = 10, so x^2 - y^2 is equal to 2/3 of x^2 + y^2. Therefore, I can multiply by 2/3 225, giving me 150. The same result, fast as well. Thanks a lot for your videos, I am an international student and the best SAT Math content to practice is yours! You are giving a third-country native the opportunity to qualify for being the best he can be in USA!

[@eeimojumo]

completed! Kind of tricky...

[@joshtaylor2592]

Good video 👍

[@cynthiachan8028]

I dont understand the third question where cube root of 16 turned into 2 times 3 root 2

[@preppros]

Hi Cynthia, are you familiar with simplifying square roots using the root tree? If so, it is very similar but since it's the third root you need to pull out a triplet instead of a pair. So 16 is the same as 2 x 2 x 2 x2 and what ever is remaining has the third root placed back in front of it, so we pull out the first 3 twos and then the last remaining two is place after the third root. So 16 is the same as 2 x 2 x 2 x2 and what ever is remaining has the third root placed back in front of it, so we pull out the first 3 twos and then the last remaining two is place after the third root.

[@mahiyemi390]

for Q5 why it can't be the answer b=5, I think the answer is, A,none

[@preppros]

Hi Mahi, No solutions, infinite solutions, and one solution are really common questions and requiring knowing what you are looking for in order for the questions to make sense. Please check out this video where I explain the concepts in more depth kzbin.info/www/bejne/bmeviqRtqspgf6M should start you at the right point but would recommend watching the rest of the video!

[@3bana]

@@prepprosI don't get why B is wrong for that question though. If x=9, then it would turn into 9x+5=9x+45

[@eeimojumo]

@@3bana I am not preppros, but for functions to have no solution, the coefficients of x and y should be the same, and the constants should be different. Also, functions with the same x and y coefficient and the same constants have infinite solutions (which means that they are the same function). So, for this problem, if the constant on the left side is equal to the constant on the right side, it would make an infinite solution, not no solution.