Results 1 to 9 of 9
Like Tree5Thanks
  • 1 Post By SlipEternal
  • 1 Post By Plato
  • 1 Post By Plato
  • 1 Post By SlipEternal
  • 1 Post By HallsofIvy

Thread: inequalities - a clarification

  1. #1
    Member
    Joined
    May 2009
    Posts
    144
    Thanks
    1

    inequalities - a clarification

    Hi everyone,

    my maths books is usually very reliable and I am always inclined to doubt myself rather than it. So, I am just checking....

    if $(x - a)^2 \geq b^2$ where a and b are constants

    then $(x - a) \geq \pm b$

    so I would say $x - a \geq b$ or $x - a \geq -b$ i.e. $x \geq a + b $ or $x \geq a - b$

    but my maths book says $x - a \leq -b$ or $x - a \geq b$ i.e. $x \leq a - b$ or $x \geq a + b$

    please tell me the book is wrong!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,530
    Thanks
    961

    Re: inequalities - a clarification

    Think about what it means. Let's use some numbers as examples. Suppose $a=0,b=5$. Then you want $x^2 \ge 5^2$. If we use your method, we would get $x \ge -5$, which would mean that $x=0$ is a solution. But $0^2 < 5^2$, does not satisfy the original inequality. You want $x \le -5$ or $x \ge 5$. The book is correct. The negative sign flips the direction of the inequality in this case.
    Thanks from s_ingram
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,152
    Thanks
    2603
    Awards
    1

    Re: inequalities - a clarification

    Quote Originally Posted by s_ingram View Post
    I am just checking....
    if $(x - a)^2 \geq b^2$ where a and b are constants
    then $(x - a) \geq \pm b$
    so I would say $x - a \geq b$ or $x - a \geq -b$ i.e. $x \geq a + b $ or $x \geq a - b$
    but my maths book says $x - a \leq -b$ or $x - a \geq b$ i.e. $x \leq a - b$ or $x \ge b$
    please tell me the book is wrong!
    $a^2\ge b^2$ if and only if $|a|\ge |b|$. Therefore, $(x-a)^2\ge b^2$ if and only if $|x-a|\ge |b|$.

    $x-a\ge |b| $ or $x-a\le -|b|$ How would you solve that?
    Thanks from SlipEternal
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    May 2009
    Posts
    144
    Thanks
    1

    Re: inequalities - a clarification

    Hi SlipEternal,

    thanks for your quick reply. I am not sure why you say that if $x \geq -5$ that x = 0 is a solution? If this were an equality, we would have no problem saying that x = +5 or x = -5 are solutions. Isn't $x \geq -5$ a solution? I understand that if you multiply both sides of an inequality by -1 it flips the inequality sign, but we are not doing this.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    May 2009
    Posts
    144
    Thanks
    1

    Re: inequalities - a clarification

    Hi Plato,

    Clearly I am missing something! I would say that the solutions would be $x \geq a + |b| $ or $x \leq a - |b|$ is something wrong with this?
    I put a = 6, b = 2 in and get $x \geq 8 $ or $x \leq 4$. I can't see any inconsistency here.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,152
    Thanks
    2603
    Awards
    1

    Re: inequalities - a clarification

    Quote Originally Posted by s_ingram View Post
    Clearly I am missing something! I would say that the solutions would be $x \geq a + |b| $ or $x \leq a - |b|$ is something wrong with this? I put a = 6, b = 2 in and get $x \geq 8 $ or $x \leq 4$. I can't see any inconsistency here.
    The original question: $(x-a)^2\ge (b)^2$

    $x-a\ge |b| $ or $x-a\le -|b|$ then it is true that $\large{x\ge a+ |b|} $ or $\large{x\le a-|b|}$
    Now lets take your example: $a=6~\&~b=2$ giving $x\ge 8\text{ or }x\le 4$

    $ \begin{align*}x&\ge 8\\\text{EX: say that }x&=10 \\|10-6|^2&\ge(2)^2\\\text{TRUE}& \end{align*}$+++++++$ \begin{align*}x&\le 4\\\text{EX: say that }x&=-1 \\|-1-6|^2&\ge(2)^2\\\text{TRUE}& \end{align*}$

    I see no inconsistency. Where is it?
    Thanks from s_ingram
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2010
    Posts
    2,530
    Thanks
    961

    Re: inequalities - a clarification

    Quote Originally Posted by s_ingram View Post
    Hi SlipEternal,

    thanks for your quick reply. I am not sure why you say that if $x \geq -5$ that x = 0 is a solution? If this were an equality, we would have no problem saying that x = +5 or x = -5 are solutions. Isn't $x \geq -5$ a solution? I understand that if you multiply both sides of an inequality by -1 it flips the inequality sign, but we are not doing this.
    $x \ge -5$ means any value for $x$ that is greater than or equal to $-5$. I chose 0 because 0 is greater than or equal to -5. Then, I plugged it into the initial inequality. In the example that I gave, you need to have $x^2 \ge 5^2$. When we plug in zero, we get $0^2 \ge 5^2$ is false. So, $x\ge -5$ is false. I just gave you a value for $x$ where $x\ge -5$, but $x^2 < 5^2$. It was an example of why $x \ge -5$ is wrong in describing $x^2\ge 5^2$.
    Thanks from s_ingram
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    May 2009
    Posts
    144
    Thanks
    1

    Re: inequalities - a clarification

    Hi Plato,

    Misunderstanding! I fully agree with everything you wrote, but you asked the question in your first post "how would you solve that?" And I thought you were pointing out an inconsistency in my arguement. In the original post I asked whether my solution or the solution in my book is correct. SlipEternal says the book is right, but I didn't understand why. Now I do, so thanks very much!
    Last edited by s_ingram; Jun 29th 2017 at 11:36 AM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,108
    Thanks
    2802

    Re: inequalities - a clarification

    Another way of looking at it: the graph of y= (x- a)^2 is a parabola, opening upward with vertex at (a, 0). The graph of y= b^2 is a horizontal straight line. (x- a)^2> b^2 is the part of the parabola above that line, two disconnected curves. The parabola passes the line where (x- a)^2= b^2 or x- a= +/- b. We don't know if b itself is positive or negative but we can write those as x= a+ |b| and a- |b|. The inequality is satisfied for x< a- |b| and x> a+ |b|.
    Thanks from s_ingram
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Dec 15th 2014, 06:31 AM
  2. Clarification?
    Posted in the Statistics Forum
    Replies: 2
    Last Post: Apr 7th 2014, 09:29 AM
  3. Clarification
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 28th 2010, 06:56 PM
  4. Clarification of PDE.
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: Feb 10th 2010, 04:35 PM
  5. At least once Clarification
    Posted in the Statistics Forum
    Replies: 6
    Last Post: Jan 7th 2010, 04:02 PM

/mathhelpforum @mathhelpforum