<p>Problem 18 on p 50 of the free SAT Prep. Booklet from CollegeBoard:
If 0<x<y and=“” (x+y)^2-(x-y)^2=“”> 25 , what is the least possible value of y?
(all < and > are underlined, meaning less or equal or greater or equal )
Grid-in
I can do this by graphing, but that took a long time, especially when you dont have ti-89
is there an easier way? like using the circle equation?</x<y></p>
<p>If you subtract the two binomials, all but their middle terms cancel, giving you:
(x+y)^2-(x-y)^2 > 25
(x^2+2xy+y^2)-(x^2-2xy+y^2) = 4xy
4xy > 25
y > 25/4x</p>
<p>the smallest value for y will occur when x is at its highest value. Since x is less than or equal to y, this is when x and y are equal.</p>
<p>therefore you can rewrite the expression
4xy > 25
to
4y^2 > 25
y^2>25/4
y>5/2
(greater than or equal to)</p>
<p>so the answer is 5/2.</p>
<p>Andre’s solution is much better then one on [url=<a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools]Collegeboard.com[/url</a>] (see under View Essays and Explanation to a free practice test).
This free test is the same as one in the CollegeBoard booklet, and it has actually been administered in March of 2005.</p>
<p>quickflood - you can answer this question without a graphing calculator.
All you need is
- to be able to sketch graphs y=x and y=1/x.
Graph of y=25/4x is just stretched out y=1/x graph, because 25/4x = (25/4)(1/x).</p>
<ol>
<li>know the following intuitive rule about graphs.</li>
</ol>
<p>A graph of y=x^2 divides the coordinate plane into two areas.
Above the graph: y>x^2 for any point (x,y),
under the graph: y<x^2 for any point (x,y).
And of course, y=x^2 on the graph itself.</p>
<p>y=x^2 is used just as an example; it could be any continuous y=f(x) instead.</p>
<p>To be continued.</p>
<p>For x>0:
y>x and y>25/4x (see Andre’s post).</p>
<p>For x>0 draw roughly y=x and y=25/4x graphs.
Shade above both y=x and y=25/4x; you’ll see a V-shaped area.
The lowest (smallest) “y” is at the “bottom” of the “V” - exactly where our two graphs intersect.
After plugging x=y into y=25/4x and simplifying,
y^2 = 25/4,
y=5/2.</p>
<p>i think that was the hardest problem in the whole practice test</p>
<p>That question was on the oct 2005 test. The way i did it was by simply taking the root of both sides. </p>
<p>(x+y)^2 -(x-y)^2 > 25
x+y-x+y >5
2y>5
y>5/2</p>
<p>Because both numbers are always positive we don’t have to worry about the negative side of the equation because you can never get 2x to equal -5 when x is positive.</p>
<p>You got lucky, icebarracuda. </p>
<p>The square root of [(x+y)^2 - (x-y)^2] is not (x+y) - (x-y). Case in point: the square root of (4^2-3^2) is not 4-3. It is 16-9, square rooted, which is the square root of 7. You need to expand the binomials and then subtract - you can’t just take the roots like that, but by chance you got lucky for this problem.</p>
<p>I was thinking the same thing, obsessed Andre.</p>
<p>Thank you so much, all of you, esp. obsessed Andre
I did try to graph this, except I didn’t know the answer would be in fraction, so I got 3.
but obsessedAndre’s way does it well
<3 thanks</p>