## A community for students. Sign up today

Here's the question you clicked on:

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

• This Question is Closed
1. dan815

m>4 right

2. Empty

Square both sides of the equation I guess

3. dan815

are u looking for int solutions?

4. anonymous

@dan815 no, real numbers solution

5. 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.

6. 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

7. anonymous

just wondering if there is any other easier method

8. 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.

9. dan815

oh so x>4 not m

10. 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.

11. anonymous

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

12. dan815

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

13. anonymous

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

14. dan815

|dw:1436389702432:dw|

15. Empty

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

16. 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 $$m \ge 1$$

17. dan815

ya thatsa the simplest way

18. anonymous

@dan815 I thought of it that way too.

19. dan815

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

20. dan815

m has to be less than that value

21. anonymous

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

22. dan815

he meant less

23. dan815

m >0 and less than 1

24. anonymous

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

25. dan815

as a slope below 0 will make it intersect behind

26. Empty

Right right sorry I am backwards in what I said

27. Empty

XD haha sorry to be confusing

28. Empty

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

29. anonymous

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

30. 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

31. anonymous

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

32. 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 0so our only solutions are where \(m^2=0$$ or $$m^2\le 1$$, which corresponds to $$|m|\le 1$$

33. 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 $$0\le m\le 1$$

34. anonymous

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

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy