How does this equation always work? $sqrt{x times (x+2)+1} = x + 1$ for non-negative $x$, and $=|x|-1$ for...












3












$begingroup$


When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:



Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$



Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $x(x+2)+1=x^2+2x+1=(x+1)^2$.
    $endgroup$
    – A.Γ.
    Jan 2 at 22:27






  • 1




    $begingroup$
    Your second formula is not true for $x=-1/2$.
    $endgroup$
    – farruhota
    Jan 2 at 22:34










  • $begingroup$
    Check your edited second equation. Are you sure it works for $-1 < x < 0$?
    $endgroup$
    – Deepak
    Jan 2 at 22:37
















3












$begingroup$


When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:



Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$



Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    $x(x+2)+1=x^2+2x+1=(x+1)^2$.
    $endgroup$
    – A.Γ.
    Jan 2 at 22:27






  • 1




    $begingroup$
    Your second formula is not true for $x=-1/2$.
    $endgroup$
    – farruhota
    Jan 2 at 22:34










  • $begingroup$
    Check your edited second equation. Are you sure it works for $-1 < x < 0$?
    $endgroup$
    – Deepak
    Jan 2 at 22:37














3












3








3





$begingroup$


When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:



Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$



Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$






algebra-precalculus







share|cite|improve this question











$endgroup$




When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:



Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$



Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 22:48









Blue

47.8k870152




47.8k870152










asked Jan 2 at 22:25









Dia. FrostDia. Frost

213




213








  • 2




    $begingroup$
    $x(x+2)+1=x^2+2x+1=(x+1)^2$.
    $endgroup$
    – A.Γ.
    Jan 2 at 22:27






  • 1




    $begingroup$
    Your second formula is not true for $x=-1/2$.
    $endgroup$
    – farruhota
    Jan 2 at 22:34










  • $begingroup$
    Check your edited second equation. Are you sure it works for $-1 < x < 0$?
    $endgroup$
    – Deepak
    Jan 2 at 22:37














  • 2




    $begingroup$
    $x(x+2)+1=x^2+2x+1=(x+1)^2$.
    $endgroup$
    – A.Γ.
    Jan 2 at 22:27






  • 1




    $begingroup$
    Your second formula is not true for $x=-1/2$.
    $endgroup$
    – farruhota
    Jan 2 at 22:34










  • $begingroup$
    Check your edited second equation. Are you sure it works for $-1 < x < 0$?
    $endgroup$
    – Deepak
    Jan 2 at 22:37








2




2




$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27




$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27




1




1




$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34




$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34












$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37




$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37










3 Answers
3






active

oldest

votes


















2












$begingroup$

Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
$$
sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
$$
You may also note that:




  • For $x geq -1$, we have $|x+1| = x+1$

  • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
    $endgroup$
    – Dia. Frost
    Jan 2 at 22:42










  • $begingroup$
    Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
    $endgroup$
    – Sambo
    Jan 2 at 23:51






  • 1




    $begingroup$
    Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
    $endgroup$
    – Sambo
    Jan 2 at 23:52



















3












$begingroup$

The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Note that you have a perfect square trinomial.



    $$x(x+2)+1 = x^2+2x+1$$



    You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



    $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



    Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



    $$x^2+2(x)(1)+1^2 = (x+1)^2$$



    So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



    $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$






    share|cite|improve this answer









    $endgroup$













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060045%2fhow-does-this-equation-always-work-sqrtx-times-x21-x-1-for-non-ne%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
      $$
      sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
      $$
      You may also note that:




      • For $x geq -1$, we have $|x+1| = x+1$

      • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


      This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
        $endgroup$
        – Dia. Frost
        Jan 2 at 22:42










      • $begingroup$
        Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
        $endgroup$
        – Sambo
        Jan 2 at 23:51






      • 1




        $begingroup$
        Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
        $endgroup$
        – Sambo
        Jan 2 at 23:52
















      2












      $begingroup$

      Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
      $$
      sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
      $$
      You may also note that:




      • For $x geq -1$, we have $|x+1| = x+1$

      • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


      This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
        $endgroup$
        – Dia. Frost
        Jan 2 at 22:42










      • $begingroup$
        Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
        $endgroup$
        – Sambo
        Jan 2 at 23:51






      • 1




        $begingroup$
        Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
        $endgroup$
        – Sambo
        Jan 2 at 23:52














      2












      2








      2





      $begingroup$

      Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
      $$
      sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
      $$
      You may also note that:




      • For $x geq -1$, we have $|x+1| = x+1$

      • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


      This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).






      share|cite|improve this answer









      $endgroup$



      Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
      $$
      sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
      $$
      You may also note that:




      • For $x geq -1$, we have $|x+1| = x+1$

      • For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)


      This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 2 at 22:35









      SamboSambo

      2,1822532




      2,1822532












      • $begingroup$
        Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
        $endgroup$
        – Dia. Frost
        Jan 2 at 22:42










      • $begingroup$
        Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
        $endgroup$
        – Sambo
        Jan 2 at 23:51






      • 1




        $begingroup$
        Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
        $endgroup$
        – Sambo
        Jan 2 at 23:52


















      • $begingroup$
        Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
        $endgroup$
        – Dia. Frost
        Jan 2 at 22:42










      • $begingroup$
        Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
        $endgroup$
        – Sambo
        Jan 2 at 23:51






      • 1




        $begingroup$
        Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
        $endgroup$
        – Sambo
        Jan 2 at 23:52
















      $begingroup$
      Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
      $endgroup$
      – Dia. Frost
      Jan 2 at 22:42




      $begingroup$
      Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
      $endgroup$
      – Dia. Frost
      Jan 2 at 22:42












      $begingroup$
      Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
      $endgroup$
      – Sambo
      Jan 2 at 23:51




      $begingroup$
      Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
      $endgroup$
      – Sambo
      Jan 2 at 23:51




      1




      1




      $begingroup$
      Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
      $endgroup$
      – Sambo
      Jan 2 at 23:52




      $begingroup$
      Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
      $endgroup$
      – Sambo
      Jan 2 at 23:52











      3












      $begingroup$

      The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$






          share|cite|improve this answer









          $endgroup$



          The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 2 at 22:27









          Dr. Sonnhard GraubnerDr. Sonnhard Graubner

          73.7k42864




          73.7k42864























              1












              $begingroup$

              Note that you have a perfect square trinomial.



              $$x(x+2)+1 = x^2+2x+1$$



              You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



              $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



              Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



              $$x^2+2(x)(1)+1^2 = (x+1)^2$$



              So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



              $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Note that you have a perfect square trinomial.



                $$x(x+2)+1 = x^2+2x+1$$



                You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



                $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



                Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



                $$x^2+2(x)(1)+1^2 = (x+1)^2$$



                So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



                $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Note that you have a perfect square trinomial.



                  $$x(x+2)+1 = x^2+2x+1$$



                  You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



                  $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



                  Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



                  $$x^2+2(x)(1)+1^2 = (x+1)^2$$



                  So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



                  $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$






                  share|cite|improve this answer









                  $endgroup$



                  Note that you have a perfect square trinomial.



                  $$x(x+2)+1 = x^2+2x+1$$



                  You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.



                  $$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$



                  Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get



                  $$x^2+2(x)(1)+1^2 = (x+1)^2$$



                  So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so



                  $$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 22:43









                  KM101KM101

                  5,9861523




                  5,9861523






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid …



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060045%2fhow-does-this-equation-always-work-sqrtx-times-x21-x-1-for-non-ne%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

                      Image
                      .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 This is a simple project showing cubes on Unity, exported to android project and imported on Android studio. On maifest file it showed following warning: Expecting <uses-feature android:name="android.software.leanback" android:required="false" /> tag I have no information about it. And also on default emptyview android project does not use/mention this feature. I know it is required by TV's and such but why its not added by default? Why is this? And Should I concern about it? android unity3d
                      Read more

                      SQL update select statement

                      Image
                      0 I'm trying to update an entry in my database (postgresql). I'm having a problem with updating the query: INSERT INTO prices (price, product_id) SELECT product_price, commn_id FROM products_temp as prod WHERE NOT EXISTS (SELECT * FROM prices od WHERE prod.commn_id = od.product_id) OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) Query works fine when i delete: OR prod.product_price != ( SELECT price FROM prices as p WHERE p.product_id = prod.commn_id order by p.end desc LIMIT 1 ) So it seems to me that it's looping through this operation. My question is, how can I fix it? sql postgresql
                      Read more

                      WPF add header to Image with URL pettitions [duplicate]

                      Image
                      up vote 0 down vote favorite This question already has an answer here: Image source URI requiring authorization 3 answers Add header to all* HTTP requests on a machine 1 answer first of all sorry for my english. Recently I added some security in the backend of my project, and now I need to send a validation code(apiKey) inside the headers to retrieve the images. I have a problem with ALL images in my project that I had attached in my UI. For example, before I had: <Image Height="90" Source="{Binding Video.Thumbnail}"/> Now I need to change my code to create a CUSTOM petition add the apiKey in the header:
                      Read more