find all value of m such that sqrt(x) = m(x-2) has a solution greater than 4

- anonymous

find all value of m such that sqrt(x) = m(x-2) has a solution greater than 4

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- dan815

m>4 right

- Empty

Square both sides of the equation I guess

- dan815

are u looking for int solutions?

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## More answers

- anonymous

@dan815 no, real numbers solution

- anonymous

@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

ok well look at the behavior of
x/(x-2)^2, find its critical, and inflection points also intersect it with y=4^2 LINE

- anonymous

just wondering if there is any other easier method

- Empty

Solve 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.

- dan815

oh so x>4 not m

- Empty

Just 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

@dan815 yeah, it's about m that makes x > 4 as solution

- dan815

you can also think about it like a sqrt function intersecting with a line displaced 2 units right, and m is the slope

- anonymous

@Empty huhm... so sqrt(4) = m(4-2), which gives m = 1. And then?

- dan815

|dw:1436389702432:dw|

- Empty

You want all solutions larger than that, so m>1. That's all your solutions.
Here look: https://www.desmos.com/calculator/ryk0xxgqmf

- Empty

The 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

ya thatsa the simplest way

- anonymous

@dan815 I thought of it that way too.

- dan815

just set x=4 and see what m has to be

- dan815

m has to be less than that value

- anonymous

@Empty well, set = m = 1.1, then x is about 3.764 which is not greater 4

- dan815

he meant less

- dan815

m >0 and less than 1

- anonymous

Yeah, as I said earlier, 0 < m < 1 but this is with the help of wolfram alpha.

- dan815

as a slope below 0 will make it intersect behind

- Empty

Right right sorry I am backwards in what I said

- Empty

XD haha sorry to be confusing

- Empty

I was imagining as m gets smaller the value of the point of their intersection increases.

- anonymous

yeah, that helps confirm our intuition but as far as the algebra go, is there a nicer way to get 0 < m < 1?

- anonymous

The 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

I'm about to go soon. So just work on it if you're interested.

- anonymous

find all value of m such that sqrt(x) = m(x-2) has a solution greater than 4
$$\sqrt{x}=m(x-2)\\x=m^2(x-2)^2\\x=m^2(x^2-2x-4)\\m^2x^2-(2m^2+1)x-4=0\\x=\frac{2m^2+1\pm\sqrt{(2m^2+1)^2+16m^2}}{2m^2}$$to prove *a* solution greater than \(4\) all we need is:$$\frac{2m^2+1+\sqrt{(2m^2+1)^2+16m^2}}{2m^2}\ge4\\2m^2+1+\sqrt{(2m^2+1)^2+16m^2}\ge8m^2\\\sqrt{(2m^2+1)^2+16m^2}\ge6m^2-1\\(2m^2+1)^2 +16m^2\ge(6m^2-1)^2\\4m^4+4m^2+1+16m^2\ge 36m^4-12m^2+1\\32m^2\ge32m^4\\m^2(m^2-1)\le 0$$so our only solutions are where \(m^2=0\) or \(m^2\le 1\), which corresponds to \(|m|\le 1\)

- anonymous

now, of course, some of these solutions are extraneous, since \(\sqrt{x}=m(x-2)\ge 0\) so \(m<0\) requires that \(x-2<0\implies x<2\) however we want solutions \(x\ge4\) we actually must have \( m>0\) so \(0\le m\le 1\)

- anonymous

now, if you wanted to guarantee \(x>4\) strictly rather than \(x\ge 4\) then the solution is really \([0,1)\)

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