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









    $endgroup$













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      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






























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                      WPF add header to Image with URL pettitions [duplicate]

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                      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:...
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