<p>This is from the Maximum SAT. Can you please show your work. Thanks! :)</p>
<ol>
<li>(page 75) If x2 (superscript)+2xy+y2(superscript) = (x-y)-2(superscript) and x+y=2 which of the following could be the value of x?</li>
</ol>
<p>a) 1/2
b) 3/5
c) 5/4
d) 2
e) 4</p>
<p>Answer is c) 5/4.</p>
<p>x+y=2
y=2-x</p>
<p>x^2+2xy+y^2 = (x-y)^-2
x^2+2xy+y^2 = 1/((x-y)^2) - negative exponent
(x+y)^2 = 1/((x-y)^2) - factor x^2 + 2xy + y^2
(x+y)^2(x-y)^2 = 1 - multiply equation by (x-y)^2
(x+2-x)^2(x-2+x)^2 = 1 - substitute
2^2(2x-2)^2 = 1 - simplify
4(2x-2)^2 = 1 - simplify
4(4x^2-8x+4) = 1 - expand
16x^2-32x+16 = 1
16x^2-32x+15 = 0</p>
<p>(32±root(1024 - 4(16)(15)))/32 - Quadratic formula
(32±root(64))/32 - simplify
(32±8)/32 - simplify
=40/32 or 24/32
=5/4 or 3/4</p>
<p>Thank you! :)</p>
<p>Also:</p>
<p>(x+y)^2(x-y)^2 = 1</p>
<p>4(x-y)^2 = 1</p>
<p>(x-y)^2 = 1/4</p>
<p>x-y = 1/2</p>
<p>Two eqns, two unknowns</p>
<p>x+y = 2
x-y = 1/2</p>
<p>add em and get</p>
<p>2x = 5/2
x = 5/4</p>
<p>x^2 + 2xy + y^2 = 1/((x-y)^2), factor LHS.</p>
<p>(x+y)^2 = 1/((x-y)^2) → 4 = 1/((x-y)^2) → (x-y)^2 = 1/4 → x-y = ± 1/2</p>
<p>So you have the system of equations</p>
<p>x+y = 2
x-y = ± 1/2</p>
<p>Add them to get 2x = 3/2 or 5/2, x = 3/4 or 5/4. 3/4 is not one of the answer choices so the answer is C) 5/4.</p>
<p>Thanks everyone! Here is another one from Maximum SAT page 125 #13.</p>
<ol>
<li> The xy-coordinate plane above shows the graph of the function y=g(x). If y=f(x), where f(x) = x^2 +1, was graphed on the same xy-coordinate plane, how many points of intersection would there be between f(x) and g(x)?</li>
</ol>
<p>Thanks again.</p>
<p>^ Cannot be solved without a picture of g(x) or the equation of g(x).</p>