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

Question

dan815 · Answer

m>4 right

empty · Answer

Square both sides of the equation I guess

dan815 · Answer

are u looking for int solutions?

anonymous · Answer

@dan815 no, real numbers solution

anonymous · Answer

@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 · Answer

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 · Answer

just wondering if there is any other easier method

empty · Answer

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 · Answer

oh so x>4 not m

empty · Answer

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 · Answer

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

dan815 · Answer

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

anonymous · Answer

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

dan815 · Answer

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