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Help on challenging math problem

  1. Jun 10, 2007 #1
    hey guys, tried these problem for awhile and cant get them. Hope i can get some help :) it would be greately aprieciated!

    the first question:
    Find all u satisfying the equation
    u+2=|7u+2|/4+u

    And the second question is:
    Find all t satisfying the inequality
    (2t+6)^1/2 >(or equal to) |t+1| -1

    I tried them both and didnt get that far and my answers seem way off, so help would be awsome. Thanks !!
     
  2. jcsd
  3. Jun 10, 2007 #2
    Please show us what you have done so far (it is standard practice on these forums not to answer homework questions, but merely to point out where people have gone wrong in the work they have done so far).
     
  4. Jun 11, 2007 #3

    VietDao29

    User Avatar
    Homework Helper

    [tex]u + 2 = \frac{|7u + 2|}{u + 4}[/tex], you mean this, right? Or, do you mean:
    [tex]u + 2 = \frac{|7u + 2|}{u} + 4[/tex]?

    Ok, assume your problem is the first one. So multiply both sides by: (u + 4) to obtain:

    [tex](u + 2) (u + 4) = |7u + 2|[/tex]

    Now, you can try to break the absolute value, i.e divide it into 2 cases 7u + 2 < 0, and 7u + 2 >= 0.

    Can you take it from here? :)

    [tex]\sqrt{2t + 6} \geq |t + 1| - 1[/tex]

    For the Square Root function to be defined, we must have 2t + 6 >= 0 ~~> t >= -3, right?

    Notice that, if we have:
    [tex]\sqrt{A} \geq B[/tex]
    Since [tex]\sqrt{A}[/tex] is always non-negative. So if B is non-negative, then the inequality will always hold, right?

    If B > 0, then you can square both sides, like this: A >= B2 and solve the inequality.

    ----------------

    Notice that in the second problem, you should divide it into 2 cases B > 0, and B <= 0. Since if B <= 0, you are not allow to square both sides, the inequality won't hold.

    You have:
    [tex]\sqrt{2} > -3[/tex], but when squaring both sides, we'll get: 2 > (-3)2 = 9, which is, of course, not true.

    Ok, can you go from here? :)
     
  5. Jun 11, 2007 #4
    hey thanks for the help, so for the first one after its broken into 2 parts
    7u+2<0 and 7u+2>=0 we would just isolate for u and get 2 solutions?

    u<7/2 and u>=7/2 ?

    and for the second one, since B<=0 wont hold and 2 isnt greater then 9 is there no real solutions?
     
  6. Jun 11, 2007 #5
    any1 got anything else that can assist me here, im struggling.
     
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