<p>Collegeboard's Pretest, Section 2 (math) #18</p>
<p>How would someone approach a problem like this?</p>
<p>If 0 ≤ x ≤ y (x+y)^2 - (x-y)^2 ≥ 25, what is the LEAST possible value of y?</p>
<p>Please put the complete question.. oh if it is then i'm sorry..</p>
<p>what is (x+y)^2 - (x-y)^2...
thats right.. its 4xy
4xy ≥ 25
I'm assuming x and y are both integers (is it?)
then 4xy has to be equal to 36 to make it the least.
x and y both can be 3... so the answer is 3..</p>
<p>Is it?</p>
<p>If it's not an integer it can be basically anything..
but to make y the least make it so that x=y
thats why i picked 36 so x can equal to y.</p>
<p>^ </p>
<p>That's what I got. I didn't post it because I thought it was wrong since the question didn't specify that it had to be an integer.</p>
<p>If it does have to be an integer, he's right.</p>
<p>Official SAT® Practice Test (a.k.a. Pretest) is the very first "new" SAT test. It was administered in March 2005.
Section 2 (math) #18 proved to be one of the most difficult questions. It was discussed A LOT on CC.</p>
<p>x and y don't have to be integers.</p>
<p>Picking up where 8parks11 got 4xy>=25.</p>
<p>There were suggested different approaches.
The most intuitive.
If we make y small, say .01, x would have to be quite big for 4xy>=25 to be true, and x<=y does not work. 4xy=25 for the lowest possible x. Let's increase y a little. Then x would decrease, but still x>y. If we keep increasing y, x would be going down, until finally x and y "meet", x=y, just as 8parks11 prophetically said, and x<=y is true at last.
4xy=25
4y^2=25
y^2=25/4
y<em>min</em>=5/2 (y is positive),</p>
<p>Straight forward approach:
25<=4xy,
25/(4y)<= x, but x<=y, so
25/(4y)<= x<= y.
25/(4y)= y
25/4 = y^2
y<em>min</em> = 5/2</p>
<p>Graphing approach.
Graph and shade two areas in the first quadrant: y>=x and y>=25/(4x). The lowest point in the common area is the point of intersection of y=x and y=25/(4x). It follows y<em>min</em>=5/2.</p>