<ol>
<li><p>We have x ~ 1/y, so 1/x ~ y, 1/(x^2) ~ y^2. You can scale everything by a constant.</p></li>
<li><p>Vieta’s formulas (or expand): 8k = m, 8+k = 5k. The second equation yields k = 2, plugging that into the first equation yields m = 16.</p></li>
</ol>
<p>I haven’t peeked at the other answers, but here goes.</p>
<h1>1. xy = k;</h1>
<p>y = k/x;
Let k = 1 (we can make the constant literally anything);
y = x^-1; - square both sides;
y^2 = x^-2;
Boom. The answer is y^2.</p>
<h1>2. is markedly harder than #1.</h1>
<p>(x-8)(x-k) = x^2 - kx - 8x + 8k;
x^2 - kx - 8x + 8k = x^2 + 5kx + m;
x^2 - (k + 8)x + 8k = x^2 + 5kx + m;</p>
<p>Therefore,
(k+8)x = 5kx;
5k = k+8;
4k = 8;
k = 2;</p>
<p>Time to plug k back into the equation to solve for m;
(x-8)(x-(2)) = x^2 - 5(2)x + m;
(x-8)(x-2) = x^2 - 10x + m;
m = -2 * -2 = 16;
m = 16;</p>