Yahoo Answers: Answers and Comments for Show that the curves y=x^42 and y=kx^2 intersect for all values of k? [Mathematics]
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From Anonymous
enUS
Thu, 10 May 2012 09:03:49 +0000
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Yahoo Answers: Answers and Comments for Show that the curves y=x^42 and y=kx^2 intersect for all values of k? [Mathematics]
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From STandardpoodle: Thanks for taking the time to answer. I hope ...
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Thu, 30 Oct 2014 06:20:50 +0000
Thanks for taking the time to answer. I hope someone can help me with my queries! I am starting my C1 revision now and having sat the paper in exam conditions have found that this is the only question I don't understand. I have three questions!
Q1) When I did it, I found the discriminant to be k^2+8 which must be above zero for all values of k. Therefore, it must have 2 roots so I thought I have shown that the curves intersect for all values of k. Why am I wrong?
Q2) I understand as far as
x = +/ sqrt(k/2 +/ sqrt(2 +k^2/4)) in the answer from Davis P above.
I understand the statement that k/2 +/ sqrt(2 + k^2/4) needs to be sown as non negative but where has the first square root gone??
Q3) My algebra skills are not strong enough to get from here k/2 +/ sqrt(2 + k^2/4) to here k/2 + sqrt(8 + k^2)/2
Can anyone help explain in detail?
Thanks

From Keith Roberts: Hi Jeff,
This is a nice problem. Okay, so we...
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Thu, 10 May 2012 09:36:18 +0000
Hi Jeff,
This is a nice problem. Okay, so we have f: y = x^4  2; and g: y = kx^2. We have to show that they intersect for all values of k. Intersect means that cross at least once, and so we are going to have to show that for each value of k, f and g intersect at least once.
Let's start by seeing where there is an intersection between f and g, if they do intersect. We do this by setting them equal:
x^4  2 = kx^2; and simplifying gives us h: x^4  kx^2  2 = 0, giving us a quadratic equation in x^2. Now we could use the quadratic formula here to actually try to generate a solution, in which you'd find that
x^2 = (k +/ sqrt(k^2 + 8))/2;
but Descartes rule of signs is probably a little more straight forward in this case, since we do not need to actually show what the solution is to h(x). Here we have
h(x) = h(x) = x^4  kx^2  2,
which has 1 sign change in the positive case and 1 sign change in the negative. Thus h will always have 1 positive real root, and 1 negative real root, and this is independent of the value of k. Therefore, we know that there are always real roots for h, meaning that f and g always intersect, regardless of the value of k.
I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below.
As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.
If we've been helpful in answering your question, please consider being one of the first to stop by our brand new Facebook page at www.facebook.com/protutorcompany.

From Davis P: equating the y's
x^4 2 = k x^2
x^4  k ...
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Thu, 10 May 2012 09:26:02 +0000
equating the y's
x^4 2 = k x^2
x^4  k x^2 2 = 0
x^4 k x^2 +(k/2)^2 = 2 + (k/2)^2
(x^2 k/2)^2 = 2 + k^2/4
x^2  k/2 = +/ sqrt( 2 + k^2/4)
x^2 = k/2 +/ sqrt(2 + k^2/4)
x = +/ sqrt(k/2 +/ sqrt(2 +k^2/4))
for the two curves to always intersect x must be real.
note all solutions for x don't have to be real only one of them
so k/2 +/ sqrt(2 + k^2/4) must be shown to be nonnegative (0 or greater)
clearly the most likely candidate for being real is to assume the + option
k/2 + sqrt(8 + k^2)/2 >=? 0
clearly this is true if k is positive.
multiply by 2
k + sqrt(8+k^2) >=? 0
the sqrt(8 + k^2) will always be > in magnitude than k so
even for negative k, k + sqrt(8 + k^2) is positive.
therefore sqrt( k/2 + sqrt(8 + k^2)/2) is real
and the two curves will intersect

From Doc W: y = kx^2 and y = x^4 2 gives
x^4  2 = kx^2...
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Thu, 10 May 2012 09:07:11 +0000
y = kx^2 and y = x^4 2 gives
x^4  2 = kx^2
x^4  kx^2  2 = 0
Let u = x^2 to give
u^2  ku  2 = 0
Completing the square gives
(u  k/2)^2  2  (k/2)^2 = 0
(u  k/2)^2 = 2 + (k/2)^2
You can see that the RHS is always > 0 so that it is ALWAYS possible to take the sqrt of BOTH side to give
u  k/2 = +/ sqrt[2 + k^2/4]
u = k/2 +/ sqrt[2 + k^2/4]
x^2 = k/2 +/ sqrt[2 + k^2/4]
and you will notice that when you take the +ve sign you get two answers for x
What happens when we take the ve sign?
You get x^2 < 0 which gives two complex solutions.
What you get are two REAL points of intersection and TWO Complex points (I wouldn't worry about these) of intersection because you have a quartic in x which has 4 solutions.
Hope this helps