Quadratic questions involving 3 variables [closed]











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If $(x-2)^2+(y-3)^2+(z+4)^2=0$ then what is the value of $x+y+z $.






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closed as off-topic by Hans Lundmark, José Carlos Santos, Chinnapparaj R, KReiser, Brahadeesh Nov 23 at 6:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hans Lundmark, José Carlos Santos, Chinnapparaj R, KReiser, Brahadeesh

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    Try adding your attempts and thoughts to the question in order to get the most appropriate answers
    – MRobinson
    Nov 22 at 10:48










  • My first thought: A square cannot be negative (for real numbers). Therefore, you are adding three terms that are all $geq 0$ ...
    – Matti P.
    Nov 22 at 10:49















up vote
-2
down vote

favorite
1












If $(x-2)^2+(y-3)^2+(z+4)^2=0$ then what is the value of $x+y+z $.






quadratics







share|cite|improve this question















closed as off-topic by Hans Lundmark, José Carlos Santos, Chinnapparaj R, KReiser, Brahadeesh Nov 23 at 6:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hans Lundmark, José Carlos Santos, Chinnapparaj R, KReiser, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 1




    Try adding your attempts and thoughts to the question in order to get the most appropriate answers
    – MRobinson
    Nov 22 at 10:48










  • My first thought: A square cannot be negative (for real numbers). Therefore, you are adding three terms that are all $geq 0$ ...
    – Matti P.
    Nov 22 at 10:49













up vote
-2
down vote

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1









up vote
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1





If $(x-2)^2+(y-3)^2+(z+4)^2=0$ then what is the value of $x+y+z $.






quadratics







share|cite|improve this question















If $(x-2)^2+(y-3)^2+(z+4)^2=0$ then what is the value of $x+y+z $.






quadratics




quadratics






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share|cite|improve this question













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edited Nov 22 at 10:48









gimusi

92.5k84495




92.5k84495










asked Nov 22 at 10:47









Hamza Afzal

2




2




closed as off-topic by Hans Lundmark, José Carlos Santos, Chinnapparaj R, KReiser, Brahadeesh Nov 23 at 6:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hans Lundmark, José Carlos Santos, Chinnapparaj R, KReiser, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Hans Lundmark, José Carlos Santos, Chinnapparaj R, KReiser, Brahadeesh Nov 23 at 6:32


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Hans Lundmark, José Carlos Santos, Chinnapparaj R, KReiser, Brahadeesh

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    Try adding your attempts and thoughts to the question in order to get the most appropriate answers
    – MRobinson
    Nov 22 at 10:48










  • My first thought: A square cannot be negative (for real numbers). Therefore, you are adding three terms that are all $geq 0$ ...
    – Matti P.
    Nov 22 at 10:49














  • 1




    Try adding your attempts and thoughts to the question in order to get the most appropriate answers
    – MRobinson
    Nov 22 at 10:48










  • My first thought: A square cannot be negative (for real numbers). Therefore, you are adding three terms that are all $geq 0$ ...
    – Matti P.
    Nov 22 at 10:49








1




1




Try adding your attempts and thoughts to the question in order to get the most appropriate answers
– MRobinson
Nov 22 at 10:48




Try adding your attempts and thoughts to the question in order to get the most appropriate answers
– MRobinson
Nov 22 at 10:48












My first thought: A square cannot be negative (for real numbers). Therefore, you are adding three terms that are all $geq 0$ ...
– Matti P.
Nov 22 at 10:49




My first thought: A square cannot be negative (for real numbers). Therefore, you are adding three terms that are all $geq 0$ ...
– Matti P.
Nov 22 at 10:49










3 Answers
3






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1
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Assuming that $x$ , $y$ and $z$ are real valued quantities, the question statement becomes valid if and only if each individual bracketed term evaluates to zero because square cannot be negative.



So



$(x-2)=0$



$(y-3)=0$



$(z+4)=0$



which gives us the solutions as $2,3$ and $-4$ respectively.






share|cite|improve this answer




























    up vote
    0
    down vote













    Hint:Note that the terms in the LHS are non-negative. Hence the only possibility of LHS to be equal to zero occurs when each of the three squares equal to zero.






    share|cite|improve this answer




























      up vote
      -1
      down vote













      HINT



      $$A^2+B^2+C^2 =0 iff A=B=C=0$$






      share|cite|improve this answer




























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        1
        down vote













        Assuming that $x$ , $y$ and $z$ are real valued quantities, the question statement becomes valid if and only if each individual bracketed term evaluates to zero because square cannot be negative.



        So



        $(x-2)=0$



        $(y-3)=0$



        $(z+4)=0$



        which gives us the solutions as $2,3$ and $-4$ respectively.






        share|cite|improve this answer

























          up vote
          1
          down vote













          Assuming that $x$ , $y$ and $z$ are real valued quantities, the question statement becomes valid if and only if each individual bracketed term evaluates to zero because square cannot be negative.



          So



          $(x-2)=0$



          $(y-3)=0$



          $(z+4)=0$



          which gives us the solutions as $2,3$ and $-4$ respectively.






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            Assuming that $x$ , $y$ and $z$ are real valued quantities, the question statement becomes valid if and only if each individual bracketed term evaluates to zero because square cannot be negative.



            So



            $(x-2)=0$



            $(y-3)=0$



            $(z+4)=0$



            which gives us the solutions as $2,3$ and $-4$ respectively.






            share|cite|improve this answer












            Assuming that $x$ , $y$ and $z$ are real valued quantities, the question statement becomes valid if and only if each individual bracketed term evaluates to zero because square cannot be negative.



            So



            $(x-2)=0$



            $(y-3)=0$



            $(z+4)=0$



            which gives us the solutions as $2,3$ and $-4$ respectively.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 at 10:55









            Akash Roy

            1




            1






















                up vote
                0
                down vote













                Hint:Note that the terms in the LHS are non-negative. Hence the only possibility of LHS to be equal to zero occurs when each of the three squares equal to zero.






                share|cite|improve this answer

























                  up vote
                  0
                  down vote













                  Hint:Note that the terms in the LHS are non-negative. Hence the only possibility of LHS to be equal to zero occurs when each of the three squares equal to zero.






                  share|cite|improve this answer























                    up vote
                    0
                    down vote










                    up vote
                    0
                    down vote









                    Hint:Note that the terms in the LHS are non-negative. Hence the only possibility of LHS to be equal to zero occurs when each of the three squares equal to zero.






                    share|cite|improve this answer












                    Hint:Note that the terms in the LHS are non-negative. Hence the only possibility of LHS to be equal to zero occurs when each of the three squares equal to zero.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 22 at 10:52









                    Thomas Shelby

                    1,170116




                    1,170116






















                        up vote
                        -1
                        down vote













                        HINT



                        $$A^2+B^2+C^2 =0 iff A=B=C=0$$






                        share|cite|improve this answer

























                          up vote
                          -1
                          down vote













                          HINT



                          $$A^2+B^2+C^2 =0 iff A=B=C=0$$






                          share|cite|improve this answer























                            up vote
                            -1
                            down vote










                            up vote
                            -1
                            down vote









                            HINT



                            $$A^2+B^2+C^2 =0 iff A=B=C=0$$






                            share|cite|improve this answer












                            HINT



                            $$A^2+B^2+C^2 =0 iff A=B=C=0$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 22 at 10:50









                            gimusi

                            92.5k84495




                            92.5k84495















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