Results 1 to 3 of 3

Math Help - Solve sevral simple inequalities with modulus and sqrts.

  1. #1
    Newbie
    Joined
    Oct 2012
    From
    Russia, Barnaul
    Posts
    2

    Post Solve sevral simple inequalities with modulus and sqrts.

    Task 1: abs(7-3x) + 4x > 5
    Task 2: sqrt(7-3x) <= x -1
    Task 3: sqrt(5-2x) > 1 - x
    Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,598
    Thanks
    1421

    Re: Solve sevral simple inequalities with modulus and sqrts.

    Quote Originally Posted by miloserdow View Post
    Task 1: abs(7-3x) + 4x > 5
    If 7- 3x< 0 (that is, if x> 7/3) this is 3x- 7+ 4x> 5. Can you solve that? If x\ge 7/3 so that 7x- 3\ge 0, this is 7- 3x+ 4x> 5. Can you solve that?

    Task 2: sqrt(7-3x) <= x -1
    \sqrt{7- 3x} is positive and if a< b, then a^2< b^2 so this is the same as [itex]7- 3x\le (x- 1)^2[/itex]. Can you solve that?

    ]Task 3: sqrt(5-2x) > 1 - x
    Same thing as task 2.

    Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3
    As in task 1, consider the cases x- 2x^2\ge 0 and x- 2x^2< 0 separately. The equation x- 2x^2= x(1- 2x)= 0 has solutions x= 0 and x= 1/2. If x= -1, x- 2x^2= -1- 2= -3< 0 so x- 2x^2< 0 for all x < 0. If x= 1/4, then x- 2x^2= 1/2- 2/16= 4/8- 1/8= 3/8> 0 so  x- 2x^2> 0 for all x between 0 and 1/2. Finally, if x= 2, x- 2x^2= 2- 8= -6< 0 so that x- 2x^2< 0 for all x> 1.

    That is, there are two cases:
    1) x< 0 or x> 1/2. In this case, x- 2x^2< 0 so |x- 2x^2|= 2x^2- x and the inequality becomes x^2*(1 + 4x^2) < 2x^2- x + 6 + 4x^3 which is the same as 4x^4+ x^2< 4x^3+ 2x^2- x+ 6 which is itself the same as 4x^4- 4x^3+ x^2+ x- 6< 0. The simplest way to solve that is to find the solutions to 4x^4- 4x^3+ x^2+ x- 6= 0 and check points between those solutions to see where they are "> 0" and where "< 0".

    2) 0< x< 1/2. In this case, x- 2x^2> 0 so |x- 2x^2|= x- 2x^2 and the inequality becomes x^2*(1 + 4x^2) < x- 2x^2 + 6 + 4x^3 which is the same as 4x^4+ x^2< 4x^3- 2x^2+ x+ 6 which is itself the same as 4x^4- 4x^3+ 3x^2- x- 6< 0.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2012
    From
    Russia, Barnaul
    Posts
    2

    Re: Solve sevral simple inequalities with modulus and sqrts.

    Thank you!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving of inequalities involving modulus
    Posted in the Algebra Forum
    Replies: 5
    Last Post: June 10th 2011, 10:28 AM
  2. Integrals - ln, sqrts
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 20th 2010, 05:13 AM
  3. Inequalities with a modulus
    Posted in the Algebra Forum
    Replies: 2
    Last Post: November 25th 2009, 01:29 AM
  4. modulus and inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: October 29th 2009, 06:25 PM
  5. Modulus and inequalities
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 23rd 2009, 08:54 PM

Search Tags


/mathhelpforum @mathhelpforum