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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

m>4 right
Square both sides of the equation I guess
are u looking for int solutions?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

@dan815 no, real numbers solution
@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.
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
just wondering if there is any other easier method
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.
oh so x>4 not m
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.
@dan815 yeah, it's about m that makes x > 4 as solution
you can also think about it like a sqrt function intersecting with a line displaced 2 units right, and m is the slope
@Empty huhm... so sqrt(4) = m(4-2), which gives m = 1. And then?
|dw:1436389702432:dw|
You want all solutions larger than that, so m>1. That's all your solutions. Here look: https://www.desmos.com/calculator/ryk0xxgqmf
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 \(m \ge 1\)
ya thatsa the simplest way
@dan815 I thought of it that way too.
just set x=4 and see what m has to be
m has to be less than that value
@Empty well, set = m = 1.1, then x is about 3.764 which is not greater 4
he meant less
m >0 and less than 1
Yeah, as I said earlier, 0 < m < 1 but this is with the help of wolfram alpha.
as a slope below 0 will make it intersect behind
Right right sorry I am backwards in what I said
XD haha sorry to be confusing
I was imagining as m gets smaller the value of the point of their intersection increases.
yeah, that helps confirm our intuition but as far as the algebra go, is there a nicer way to get 0 < m < 1?
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
I'm about to go soon. So just work on it if you're interested.
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 \(|m|\le 1\)
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 \(0\le m\le 1\)
now, if you wanted to guarantee \(x>4\) strictly rather than \(x\ge 4\) then the solution is really \([0,1)\)

Not the answer you are looking for?

Search for more explanations.

Ask your own question