Yes, this is an easy problem. I was able to solve it in my head. The graphs intersect in one point when k = 0 or 4 and in two points when k is negative of greater then 4. No intersecting happens when k is 1, 2 or 3. For k ≥ 0 we need to fulfil x/2 < √x in order to make the graph not intersect, because x + y = k is a line with slope of -1 and xy = k is hyperbole in first and third quadrant with the x and y axes being the asymptotes. So, we can compare the point on the line y = x at which these graphs intersect it and that's were we get the x/2 < √x from. Since √5 < 5/2 and the derivate of x/2 is constant 1/2 while the derivative of √x is 1/(2√5) < 1/2 at x=5 and goes to zero when x goes to infinity, if is clear x/2 > √x for x ≥ 5. We only need to check k = 0, 1, 2, 3 or 4. Clearly k = 0 or k = 4 don't work leaving us k = 1, 2 or 3. For k < 0, x + y = k is still a line with slope of -1, but xy = k becomes hyperbole in the second and fourth quadrants intersecting the line always once in both of these quadrants.
@@electricgamer_yt4753 k=2 is an integer, is it not? I was providing an example of a pair of complex numbers that when added or multiplied give the same integer result.
Poor doubt. Once think what is difference between argand plane and cartesian plane. One have it's axis as imaginary and real,while later have as x and y
Yes, this is an easy problem. I was able to solve it in my head. The graphs intersect in one point when k = 0 or 4 and in two points when k is negative of greater then 4. No intersecting happens when k is 1, 2 or 3.
For k ≥ 0 we need to fulfil x/2 < √x in order to make the graph not intersect, because x + y = k is a line with slope of -1 and xy = k is hyperbole in first and third quadrant with the x and y axes being the asymptotes. So, we can compare the point on the line y = x at which these graphs intersect it and that's were we get the x/2 < √x from. Since √5 < 5/2 and the derivate of x/2 is constant 1/2 while the derivative of √x is 1/(2√5) < 1/2 at x=5 and goes to zero when x goes to infinity, if is clear x/2 > √x for x ≥ 5. We only need to check k = 0, 1, 2, 3 or 4. Clearly k = 0 or k = 4 don't work leaving us k = 1, 2 or 3.
For k < 0, x + y = k is still a line with slope of -1, but xy = k becomes hyperbole in the second and fourth quadrants intersecting the line always once in both of these quadrants.
Very nice! And I'm not just commenting about the shirt.
😂
Thanks for adding the graph. Maybe it would also help to visualize the problem early on by making a few hand drawn graphs on the blackboard.
The hat's nice, but that's sum shirt you're wearing; it's even cooler.
How about the complex numbers (1+i) and (1-i)? (1+i)+(1-i)=2 and (1+i)(1-i)=2
It says (k€Z)
@@electricgamer_yt4753 k=2 is an integer, is it not? I was providing an example of a pair of complex numbers that when added or multiplied give the same integer result.
I had the same question but the question referred to the graphs, so it clearly requires x and y to be real.
@@PrimeNewtons I understand that, my point was that there are pairs of complex numbers whose sum and product is equal and is an integer.
Poor doubt.
Once think what is difference between argand plane and cartesian plane.
One have it's axis as imaginary and real,while later have as x and y
I also solved it like this too
My fav title.
First time I saw a Prime Newtons video and I could solve it instantly lol
Those found k values aren't for the case where the 2 graphs ARE intersecting, while the question is about NOT intersecting ????
In 6:25 he shows us that between 0 and 4 they are not intersecting, so those are good answers
What's wrong with the audio?
I was surprised too. There was an update just before I started recording. I'll check it out.
yeah lol i was thinking i stumbled upon an older video
Audio is messed up compared to other videos…
Can cannot?
Although it may sound strange to the ear but it actually makes sense. What the title basically says is ''what values can not to be k?".