<p>On barronstestprep.com, I came across this question:</p>
<p>If x is negative, which of the following must be true?</p>
<p>I. x^3 < x^2
II. x + (1/x)< 0
III. x = square root of (x^2)</p>
<p>A I only
B II only
C I and II only
D II and III only
E I, II, and III</p>
<p>I chose E because (I.) a negative number raised to the third is negative and so is less than that number squared, (II.) a negative plus a negative is a negative and so is less than zero, and (III.) because any square root must be denoted as positive or negative and so the square root could be negative and equal to x.</p>
<p>Barron, however, said this:</p>
<p> To solve this problem, examine each statement independently to determine if it is true or false.
I. If x is negative, then x^3 is negative and so must be less than x^2, which is positive. (I is true.)
II. If x is negative, so is, (1/x) and the sum of two negative numbers is negative. (II is true.)
III. The square root of a number is never negative, and so could not possibly equal x. (III is false.)
Only I and II are true. The answer is C.</p>
<p>Is this correct? (because this goes against what I learn in school) If it isn't/is does anyone know how they would handle square roots on the SAT? (ie. if this was a question on the real SAT, do they stand by the fact that square roots cannot be negative?)</p>
<p>bluetortoise this is not an imaginary number. An imaginary number is the sqroot of a negative. If you take the sqroot of something squared you don’t get an imaginary. Sqroot of (-3)^2 is 3. However, the issue here is that I think they have to specify that the result is +/-sqroot for a negative result to be possible, and I have never seen them do this. Even if it was +/-sqroot x^2 III would not necessarily be true because it could either equal -x or x. I learned this the hard way by the way…</p>
<p>The key word here is “must”. It mean it absolutely has to be true. Consider x=-2. X^2=4. Now the square root of 4 could be either 2 or -2. So III is false becuz there is an option which says square root of x is >0. Watch out for statements that say “must be” and “could be”. One word but they change the entire question</p>
<p>I seem to have misread the condition. However, concept is still the same. Underoot 4 could be either 2 or -2. And taking X as -2, 2 does not equal -2</p>
<p>This is a common source of confusion. </p>
<p>First, the symbol square root is used to denote the non negative square root of the non negative number n. Therefore, square root(read the symbol here) of 16 is 4. If we write, -square root(16), then that is equal to -4. Therefore, the symbol only represents one value. Square root of 16 is not equal to -4. </p>
<p>In the above example, if we substitute x=-3, in the relationship:
III. x = square root(x^2)
then we have Is -3 = square root([-3]^2)
Now [-3]^2 = 9, however, square root of 9 is always 3, because that is what the symbol denotes. So in the above example, III is not true for negative values of x. </p>
<p>And yes, this is how the SAT treats the symbol of square root.</p>
<p>^ decent level math requires for a square root to be equal to +,- watever value has been obtained</p>
<p>“And yes, this is how the SAT treats the symbol of square root.”</p>
<p>Can this be confirmed? Is it in the blue book or anything?</p>
<p>I’m pretty sure the reason why III is out is because the square root of any number squared gives two results.</p>
<p>So like with -n (where n is any integer), we would have (-n)^2 = n^2, then the square root of that gives both -n AND +n. Since we have two different possible answers, then III can be true or it can be false, but not must.
This is what a couple of people have already stated.</p>
<p>Also, there are multiple ways to arrive at the same conclusion in math. Pick a method that works best for you.</p>
<p>@ybrown234</p>
<p>Please see the image below for an extract from a collegeboard document that clarifies this:</p>
<p>[SATprincipalsquareroot</a> - Dabral’s library](<a href=“http://www.screencast.com/t/f6fKeIBXo]SATprincipalsquareroot”>http://www.screencast.com/t/f6fKeIBXo)</p>