<p>Let f(x)=x^2 - 5x +2 and g(x) = f(x-4)
Find the positive root of the equition f(x) = g(2)</p>
<p>I solved this ,the answer is 7, but do you thnik problem like this appears on the SAT ?It is from Barron`s 2400 club and is marked as ‘‘hard’’ In this book there are no ‘easy’ problems and the medium ones are the ones that we consider as ‘‘hard’’ on the real test.So this is supposed to be a ‘‘superhard’’ :}</p>
<p>i don’t think it will be on it.
I’ve seen problems like this and this is way above the level of an ETS question.</p>
<p>This type of question is likely to appear on the SAT IIC but not on SAT I Math.Though the question itself isn’t that difficult, it is possible this could be on the Last few on…</p>
<p>I love books like Barron`a 2400 that prepare you for the hardest things and you feel the real test much easier.I recommend it it.It is great and somehow while reading it,you become a friend with the author and understand everything he/she says</p>
<p>That’s not an SAT I question; it’s too straightforward, lol.</p>
<p>^agreed</p>
<p>it’s likely to show up as an easy level problem on the SAT II math, but for SAT I, it seems too straightforward and also because it wayyy too time consuming. I dont see how you could solve it in <30 seconds, which is the amount of time each math problem on SAT I is designed to take you…that is, if you do it the ETS way.</p>
<p>In <30 sec:
g(2) = f(2-4)
g(2) = f(-2)
f(x) = f(-2)
Since x=2.5 is an axis of symmetry of parabola y=x^2-5x+2 (remember -b/2a)
f(7) = f(-2)</p>
<p>Wow you don`t have 30 sec only,you have about a minute .For example 18 questions for 20 minutes;]
But you may spend 1.5 minutes for this type of questions and no more than 20 seconds f or each of the first 3 questions</p>
<p>If you’re (really) good, all problems should take the same time; that is, they should all be similarly easy.</p>
<p>Ok ,so according to , 2x+2 = 8 and the above problem take the same amount of time ?</p>
<p>Well, the real SAT doesn’t have computationally intensive problems; higher difficulty problems just have “trickier” solutions.</p>