<p>how do i do this?</p>
<p>It has got to be A.</p>
<p>2 can’t make 1 even because it is too small to be that big of an integer to be K.</p>
<p>It is the only answer, which can satisfy the graph as if you plug it into the calculator, it gives you a small opened circle as the turning point.</p>
<p>Not sure if I’m accurate, but other CCers can come here to agree/disagree with me.</p>
<p>To do this, you plug this into your calculator and graph. You substitute the k-values in and find where the graphs intersect. If the x-coordinate (of the point of intersection) is not an integer, then that’s the right k-value. I suppose you could set x^2 equal to k-x^2, but I would think that that’s more time consuming and error-prone.</p>
<p>Line segment AB would be equal to the x value of point B, which is the intersection of the two graphs</p>
<p>y = x^2
y = k - x^2</p>
<p>Equal them to each other to find the intersection point</p>
<p>x^2 = k - x^2
2x^2 = k
x^2 = k/2
x = sqrt(k/2)</p>
<p>Now, substitute all values of k from the options and see which one isn’t an integer.</p>
<p>You’ll notice that 12 is the only one that doesn’t give you an integer as sqrt(12/2) = sqrt(6). The answer is C.</p>
<p>WHOA! Bassir made it very simple. Thanks</p>