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anonymous
 one year ago
find all value of m such that sqrt(x) = m(x2) has a solution greater than 4
anonymous
 one year ago
find all value of m such that sqrt(x) = m(x2) has a solution greater than 4

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Empty
 one year ago
Best ResponseYou've already chosen the best response.1Square both sides of the equation I guess

dan815
 one year ago
Best ResponseYou've already chosen the best response.1are u looking for int solutions?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 no, real numbers solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Empty I guess you could do so. I've attempted and with a help with wolfram alpha, it got 0 < m < 1. But the arithmetic is just way tedious.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ok well look at the behavior of x/(x2)^2, find its critical, and inflection points also intersect it with y=4^2 LINE

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just wondering if there is any other easier method

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Solve for x, then set that value to be greater than 4. Hmmm Ohhh so you're looking for a trick, like a nice way to do it ok that makes this much more fun hahah ok hmmm give me a minute.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Just plug in x=4 and solve for m=1. Then you know m has to be greater than that, cause the square root is always positive, done.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 yeah, it's about m that makes x > 4 as solution

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you can also think about it like a sqrt function intersecting with a line displaced 2 units right, and m is the slope

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Empty huhm... so sqrt(4) = m(42), which gives m = 1. And then?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1You want all solutions larger than that, so m>1. That's all your solutions. Here look: https://www.desmos.com/calculator/ryk0xxgqmf

Empty
 one year ago
Best ResponseYou've already chosen the best response.1The case you solved for is the lowest possible case for the value of m. Well technically it's not because it's really m>1 not \(m \ge 1\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ya thatsa the simplest way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 I thought of it that way too.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1just set x=4 and see what m has to be

dan815
 one year ago
Best ResponseYou've already chosen the best response.1m has to be less than that value

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Empty well, set = m = 1.1, then x is about 3.764 which is not greater 4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, as I said earlier, 0 < m < 1 but this is with the help of wolfram alpha.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1as a slope below 0 will make it intersect behind

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Right right sorry I am backwards in what I said

Empty
 one year ago
Best ResponseYou've already chosen the best response.1XD haha sorry to be confusing

Empty
 one year ago
Best ResponseYou've already chosen the best response.1I was imagining as m gets smaller the value of the point of their intersection increases.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, that helps confirm our intuition but as far as the algebra go, is there a nicer way to get 0 < m < 1?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The tedious algebra is solving this, which wolfram alpha did https://www.wolframalpha.com/input/?i=%28%284m%5E2%2B1%29+%2B+sqrt%5B+%284m%5E2%2B1%29%5E2++4%28m%5E2%29%284m%5E2%29%5D+%29+%2F+%282m%5E2%29+%3E+4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm about to go soon. So just work on it if you're interested.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0find all value of m such that sqrt(x) = m(x2) has a solution greater than 4 $$\sqrt{x}=m(x2)\\x=m^2(x2)^2\\x=m^2(x^22x4)\\m^2x^2(2m^2+1)x4=0\\x=\frac{2m^2+1\pm\sqrt{(2m^2+1)^2+16m^2}}{2m^2}$$to prove *a* solution greater than \(4\) all we need is:$$\frac{2m^2+1+\sqrt{(2m^2+1)^2+16m^2}}{2m^2}\ge4\\2m^2+1+\sqrt{(2m^2+1)^2+16m^2}\ge8m^2\\\sqrt{(2m^2+1)^2+16m^2}\ge6m^21\\(2m^2+1)^2 +16m^2\ge(6m^21)^2\\4m^4+4m^2+1+16m^2\ge 36m^412m^2+1\\32m^2\ge32m^4\\m^2(m^21)\le 0$$so our only solutions are where \(m^2=0\) or \(m^2\le 1\), which corresponds to \(m\le 1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, of course, some of these solutions are extraneous, since \(\sqrt{x}=m(x2)\ge 0\) so \(m<0\) requires that \(x2<0\implies x<2\) however we want solutions \(x\ge4\) we actually must have \( m>0\) so \(0\le m\le 1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now, if you wanted to guarantee \(x>4\) strictly rather than \(x\ge 4\) then the solution is really \([0,1)\)
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