<p>There is a question on the December 2005 ACT test that puzzles me....
How many (x,y) pairs of real numbers satisfy xy=3 and (x+y)(squared)=10?
a.0 b.1 c.2 d.4 e.infinitely many</p>
<p>Why is the answer 0?
And are the old ACT tests harder than the ones now?</p>
<p>From what your post said… The only factors of 3 are 3 and 1… and (3+1)^2 does not equal ten. Actually NOTHING squared equals 10. Idk, maybe I’m missing something.</p>
<p>If you plot this out on your graphing calculator:</p>
<p>y1 = 3/x
y2 = sqrt(10)-x (to get this, take the sqrt of both sides and subtract x)</p>
<p>There are ZERO intersection points between these two curves.</p>
<p>without a graphing calc:
y = 3/x
y = 10^.5 - x</p>
<p>set them equal 3/x = sqrt(10) - x</p>
<p>multiply both sides by ‘x’:</p>
<p>3 = sqrt(10)x - x^2, rearrange to:</p>
<p>x^2 - sqrt(10)x + 3 = 0, use the quadratic formula, and you will get no real answers. Even if 0 somehow came up, it would never work as it would be undefined, just a caution measure if it did seeing that if you multiplied both sides by x^2, you would get </p>
<p>3x + 3x^3 = sqrt(10)x^2, which says x =0 would work, but actually doesn’t in the real equation, which is really interesting imo.</p>