![\"new](\"https:\/\/s3.amazonaws.com\/magoosh-company-site\/wp-content\/uploads\/hs\/files\/2016\/03\/10021507\/screen-shot-2016-02-24-at-9.43.36-am.png\")
<\/p>\nyou may have heard that the new sat math section is less tricky and more straightforward than its old sat counterpart. yes, it is true that there are more way more algebra problems, and the wording of the questions are closer to what you would see on a typical math test in school. <\/p>\n
however, don’t be fooled into thinking that there’s only one way to solve each problem. there are many occasions where the test makers will present you with a question that can be solved in a couple of different ways. usually, one method is faster but less obvious than the other method.<\/p>\n
if you want to learn how to save yourself precious seconds on the test so you can get a better score, read on.
\n <\/p>\n
here at magoosh, we love to throw out an sat math challenge question here or there to tickle your brain and get you ready for the real deal. you may have seen this question floating around the border of our blog. take a look at the problem below and think about how you would approach it.<\/p>\n
<\/p>\n
<\/p>\n
when you first see the problem, you might have the strong urge to just use brute force math and solve for x mathematically. after all, this is pretty basic algebra.<\/p>\n
1. square both sides<\/strong><\/p>\n this gets rid of the square root on the right side of the equation.<\/p>\n 2. foil the left side<\/strong><\/p>\n this breaks apart the right side of the equation so that we can get ready for the third step.<\/p>\n 3. make the right side equal to 0<\/strong><\/p>\n by doing this, we can figure out the roots to our equation.<\/p>\n 4. factor<\/strong><\/p>\n we find out that we have two answers: x = \u2013 1 and x = 15. however, plugging in x = \u2013 1 into the original equation yields \u2013 3 = 3, so our answer is just x = 15. 1. notice the radical<\/strong><\/p>\n if we slow down a bit and recognize from the very beginning that the right side of the equation has a radical covering it, we know that that side has to be positive. otherwise, taking the square root of a negative number yields an imaginary number.<\/p>\n 2. both sides of the equation must be positive<\/strong><\/p>\n therefore, we know that the left side of the equation also has to be positive or imaginary. looking at our answer choices, we see that there are no imaginary numbers, so both sides of the equation have to be positive.<\/p>\n 3. work backwards<\/strong><\/p>\n working backwards, the only possible answer from the answer choices we can plug in for x that yields a positive number is 15. sometimes, the challenge lies in being able to figure out how to solve a problem the smart way. you only have a limited amount of time to devote to each question, so don’t overthink it. <\/p>\n the next time you come across a tough sat math challenge question, try to find the quick method! the new sat math section is less tricky and more straightforward than the old sat. but this doesn’t mean there aren’t tricks to solving challenging problems. <\/p>\n","protected":false},"author":158,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[91],"tags":[],"ppma_author":[24918],"acf":[],"yoast_head":"\n
\n <\/p>\nsat math challenge question: the quick method<\/h2>\n
\n <\/p>\ntaking it a step further<\/h2>\n
\n <\/p>\n","protected":false},"excerpt":{"rendered":"