When to apply negative sign when number is squared












2












$begingroup$


I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?






algebra-precalculus recreational-mathematics







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  • 2




    $begingroup$
    because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
    $endgroup$
    – Luke
    47 mins ago










  • $begingroup$
    Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
    $endgroup$
    – Henry
    34 mins ago


















2












$begingroup$


I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?






algebra-precalculus recreational-mathematics







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
    $endgroup$
    – Luke
    47 mins ago










  • $begingroup$
    Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
    $endgroup$
    – Henry
    34 mins ago
















2












2








2





$begingroup$


I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?






algebra-precalculus recreational-mathematics







share|cite|improve this question









$endgroup$




I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?






algebra-precalculus recreational-mathematics




algebra-precalculus recreational-mathematics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked 51 mins ago









JohnJohnyPapaJohnJohnJohnyPapaJohn

606




606








  • 2




    $begingroup$
    because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
    $endgroup$
    – Luke
    47 mins ago










  • $begingroup$
    Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
    $endgroup$
    – Henry
    34 mins ago
















  • 2




    $begingroup$
    because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
    $endgroup$
    – Luke
    47 mins ago










  • $begingroup$
    Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
    $endgroup$
    – Henry
    34 mins ago










2




2




$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
47 mins ago




$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
47 mins ago












$begingroup$
Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
34 mins ago






$begingroup$
Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
34 mins ago












3 Answers
3






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$begingroup$

When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    As it is already in the previous answers:
    $(-x)^2neq-x^2$
    To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$






    share|cite|improve this answer








    New contributor




    user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$





















      -1












      $begingroup$

      Does this question arise from the task to complete a square?



      Like $x^2+x=(x+1/2)^2-(1/2)^2$?



      In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.



      For $x^2+bx+c$ this $a$ would be given by $left(frac{b}2right)^2$. And so $-a=-left(frac{b}2right)^2$



      And because we want to have this zero, the sign can not be influenced by the square.
      Else it would not add up to be $0$.






      share|cite|improve this answer









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      • $begingroup$
        And the downvote is for what?
        $endgroup$
        – Cornman
        28 mins ago












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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

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      active

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      active

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      4












      $begingroup$

      When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)






          share|cite|improve this answer









          $endgroup$



          When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 48 mins ago









          Minus One-TwelfthMinus One-Twelfth

          3,603413




          3,603413























              0












              $begingroup$

              As it is already in the previous answers:
              $(-x)^2neq-x^2$
              To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$






              share|cite|improve this answer








              New contributor




              user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$


















                0












                $begingroup$

                As it is already in the previous answers:
                $(-x)^2neq-x^2$
                To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$






                share|cite|improve this answer








                New contributor




                user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  As it is already in the previous answers:
                  $(-x)^2neq-x^2$
                  To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$






                  share|cite|improve this answer








                  New contributor




                  user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$



                  As it is already in the previous answers:
                  $(-x)^2neq-x^2$
                  To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$







                  share|cite|improve this answer








                  New contributor




                  user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered 19 mins ago









                  user665960user665960

                  113




                  113




                  New contributor




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                  New contributor





                  user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                  user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.























                      -1












                      $begingroup$

                      Does this question arise from the task to complete a square?



                      Like $x^2+x=(x+1/2)^2-(1/2)^2$?



                      In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.



                      For $x^2+bx+c$ this $a$ would be given by $left(frac{b}2right)^2$. And so $-a=-left(frac{b}2right)^2$



                      And because we want to have this zero, the sign can not be influenced by the square.
                      Else it would not add up to be $0$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        And the downvote is for what?
                        $endgroup$
                        – Cornman
                        28 mins ago
















                      -1












                      $begingroup$

                      Does this question arise from the task to complete a square?



                      Like $x^2+x=(x+1/2)^2-(1/2)^2$?



                      In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.



                      For $x^2+bx+c$ this $a$ would be given by $left(frac{b}2right)^2$. And so $-a=-left(frac{b}2right)^2$



                      And because we want to have this zero, the sign can not be influenced by the square.
                      Else it would not add up to be $0$.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        And the downvote is for what?
                        $endgroup$
                        – Cornman
                        28 mins ago














                      -1












                      -1








                      -1





                      $begingroup$

                      Does this question arise from the task to complete a square?



                      Like $x^2+x=(x+1/2)^2-(1/2)^2$?



                      In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.



                      For $x^2+bx+c$ this $a$ would be given by $left(frac{b}2right)^2$. And so $-a=-left(frac{b}2right)^2$



                      And because we want to have this zero, the sign can not be influenced by the square.
                      Else it would not add up to be $0$.






                      share|cite|improve this answer









                      $endgroup$



                      Does this question arise from the task to complete a square?



                      Like $x^2+x=(x+1/2)^2-(1/2)^2$?



                      In this case you want to use a binomial formula. For that you add a zero, by writing $0=a-a$. This $a$ happens to be a square.



                      For $x^2+bx+c$ this $a$ would be given by $left(frac{b}2right)^2$. And so $-a=-left(frac{b}2right)^2$



                      And because we want to have this zero, the sign can not be influenced by the square.
                      Else it would not add up to be $0$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 41 mins ago









                      CornmanCornman

                      3,82421233




                      3,82421233












                      • $begingroup$
                        And the downvote is for what?
                        $endgroup$
                        – Cornman
                        28 mins ago


















                      • $begingroup$
                        And the downvote is for what?
                        $endgroup$
                        – Cornman
                        28 mins ago
















                      $begingroup$
                      And the downvote is for what?
                      $endgroup$
                      – Cornman
                      28 mins ago




                      $begingroup$
                      And the downvote is for what?
                      $endgroup$
                      – Cornman
                      28 mins ago


















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