<p>if 0 ≤ x ≤ y and (x+y)^2 - (x-y)^2 ≥ 25 =, what is the least possible value of y?</p>
<p>just kidding. the least possible value for y is 4.</p>
<p>how do you figure 4?</p>
<p>Do x and y have to be integers? If so, I agree with Chone, 4.</p>
<p>well i just figured the problem out and its not 4. it does not say that x and y are or have to be integers so its safe to say it can be a decimal or fraction</p>
<p>(x+y)^2 - (x-y)^2 ≥ 25
4xy >= 25
xy >= 25/4
y >= x -------> y^2 >= xy
y^2 >= xy >= 25/4
y^2 >= 25/4
y >= 5/2
y min = 5/2,
x max = 5/2</p>
<p>you got it ! but i was wondering on your second step how did it become 4xy?</p>
<p>multiply out - </p>
<p>(x+y)^2 = x^2+2xy+y^2
(x-y)^2 = x^2-2xy+y^2</p>
<p>x^2+2xy+y^2-(x^2-2xy+y^2) = x^2+2xy+y^2-x^2+2xy-y^2=4xy</p>
<p>A simpler way of getting 4xy:</p>
<p>(x+y)^2 - (x-y)^2
(x+y+x-y)(x+y-x+y)------>(a^2 - b^2)=(a+b)(a-b)
(2x)(2y)
4xy</p>
<p>There ya go :)</p>
<p>I got 2.5 first and was wondering how people got 4. lol</p>
<p>I got it firster. :D
<a href="http://talk.collegeconfidential.com/sat-preparation/405228-math-problem-help.html%5B/url%5D">http://talk.collegeconfidential.com/sat-preparation/405228-math-problem-help.html</a></p>
<p>(x+y+x-y)*(x+y-x+y)=25
4xy=25</p>
<p>Then the answer is 4</p>
<p>^ How is it 4?</p>