Let $a in mathbb R$. Prove that ($x^2 + ax + a > 0$ for every $x in mathbb R)$ iff $(0< a<4)$...












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This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.



Thanks,
John






proof-writing quadratics







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closed as off-topic by Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos Jan 25 at 13:09


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  • $begingroup$
    So, what happens when you do complete the square?
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    – Lord Shark the Unknown
    Jan 25 at 4:51
















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$begingroup$



  1. List item


This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.



Thanks,
John






proof-writing quadratics







share|cite|improve this question











$endgroup$



closed as off-topic by Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos Jan 25 at 13:09


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos

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
















  • $begingroup$
    So, what happens when you do complete the square?
    $endgroup$
    – Lord Shark the Unknown
    Jan 25 at 4:51














-1












-1








-1





$begingroup$



  1. List item


This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.



Thanks,
John






proof-writing quadratics







share|cite|improve this question











$endgroup$





  1. List item


This is a homework question for my university math proofs course. I am asked to prove the above bi conditional statement. The text book gives the hint that completing the square of the quadratic equation will simplify it.



Thanks,
John






proof-writing quadratics




proof-writing quadratics






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edited Jan 25 at 5:00









Andrei

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asked Jan 25 at 4:50









John TophamJohn Topham

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closed as off-topic by Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos Jan 25 at 13:09


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos

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







closed as off-topic by Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos Jan 25 at 13:09


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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord Shark the Unknown, T. Bongers, metamorphy, Cesareo, José Carlos Santos

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












  • $begingroup$
    So, what happens when you do complete the square?
    $endgroup$
    – Lord Shark the Unknown
    Jan 25 at 4:51


















  • $begingroup$
    So, what happens when you do complete the square?
    $endgroup$
    – Lord Shark the Unknown
    Jan 25 at 4:51
















$begingroup$
So, what happens when you do complete the square?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:51




$begingroup$
So, what happens when you do complete the square?
$endgroup$
– Lord Shark the Unknown
Jan 25 at 4:51










5 Answers
5






active

oldest

votes


















0












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If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:



$$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$



The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This answer is easiest for me to understand thanks
    $endgroup$
    – John Topham
    Jan 25 at 5:11



















0












$begingroup$

Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let $x^2+ax+a=y$



    $iff x^2+ax+a-y=0$



    As $x$ is real, the discriminant must be $ge0$



    $a^2-4(a-y)ge0iff4yge a(4-a)$



    It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Complete the square:
      $$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
      left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$






      share|cite|improve this answer









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        0












        $begingroup$

        After completing the square of the above equation you will get:
        $$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
        which is nothing but:
        $$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
        The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
        Now for the second part
        $$a-frac{a^2}{4}gt0$$
        which gives $$4a-a^2gt0$$
        Or $$a^2-4alt0$$
        $$a(a-4)lt0$$ which gives the critical points $a=0,4$
        Now this inequality only holds if $0lt alt4$



        Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
        One more thing is you can check for the vertex of the parabola....
        Hope this helps






        share|cite|improve this answer









        $endgroup$




















          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:



          $$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$



          The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This answer is easiest for me to understand thanks
            $endgroup$
            – John Topham
            Jan 25 at 5:11
















          0












          $begingroup$

          If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:



          $$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$



          The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This answer is easiest for me to understand thanks
            $endgroup$
            – John Topham
            Jan 25 at 5:11














          0












          0








          0





          $begingroup$

          If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:



          $$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$



          The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.






          share|cite|improve this answer









          $endgroup$



          If you want to prove a quadratic is $>0$, I'd always recommend Completing the Square. We have:



          $$x^2+ax+a=(x-frac a2)^2-frac{a^2}{4}+a$$



          The squared part is obviously greater than $0$, and the other term is greater than $0$ whenever $4a>a^2$, this is when $ain(0,4)$, as stated in the question.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 25 at 5:05









          Rhys HughesRhys Hughes

          7,0401630




          7,0401630












          • $begingroup$
            This answer is easiest for me to understand thanks
            $endgroup$
            – John Topham
            Jan 25 at 5:11


















          • $begingroup$
            This answer is easiest for me to understand thanks
            $endgroup$
            – John Topham
            Jan 25 at 5:11
















          $begingroup$
          This answer is easiest for me to understand thanks
          $endgroup$
          – John Topham
          Jan 25 at 5:11




          $begingroup$
          This answer is easiest for me to understand thanks
          $endgroup$
          – John Topham
          Jan 25 at 5:11











          0












          $begingroup$

          Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.






          share|cite|improve this answer









          $endgroup$


















            0












            $begingroup$

            Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.






            share|cite|improve this answer









            $endgroup$
















              0












              0








              0





              $begingroup$

              Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.






              share|cite|improve this answer









              $endgroup$



              Suppose $x^2+ax + a > 0$ which is to say that no real roots can exist and so $a^2-4a < 0 $ and notice that $a^2-4a$ is a parabola and it's below the x-axis when $0<a<4$ as one can check easily. The other direction is done similarly.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 25 at 4:54









              Jimmy SabaterJimmy Sabater

              3,023325




              3,023325























                  0












                  $begingroup$

                  Let $x^2+ax+a=y$



                  $iff x^2+ax+a-y=0$



                  As $x$ is real, the discriminant must be $ge0$



                  $a^2-4(a-y)ge0iff4yge a(4-a)$



                  It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$






                  share|cite|improve this answer









                  $endgroup$


















                    0












                    $begingroup$

                    Let $x^2+ax+a=y$



                    $iff x^2+ax+a-y=0$



                    As $x$ is real, the discriminant must be $ge0$



                    $a^2-4(a-y)ge0iff4yge a(4-a)$



                    It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$






                    share|cite|improve this answer









                    $endgroup$
















                      0












                      0








                      0





                      $begingroup$

                      Let $x^2+ax+a=y$



                      $iff x^2+ax+a-y=0$



                      As $x$ is real, the discriminant must be $ge0$



                      $a^2-4(a-y)ge0iff4yge a(4-a)$



                      It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$






                      share|cite|improve this answer









                      $endgroup$



                      Let $x^2+ax+a=y$



                      $iff x^2+ax+a-y=0$



                      As $x$ is real, the discriminant must be $ge0$



                      $a^2-4(a-y)ge0iff4yge a(4-a)$



                      It is sufficient to have $a(4-a)>0iff a(a-4)<0iff0<a<4$







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 25 at 4:58









                      lab bhattacharjeelab bhattacharjee

                      227k15158275




                      227k15158275























                          0












                          $begingroup$

                          Complete the square:
                          $$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
                          left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$






                          share|cite|improve this answer









                          $endgroup$


















                            0












                            $begingroup$

                            Complete the square:
                            $$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
                            left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$






                            share|cite|improve this answer









                            $endgroup$
















                              0












                              0








                              0





                              $begingroup$

                              Complete the square:
                              $$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
                              left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$






                              share|cite|improve this answer









                              $endgroup$



                              Complete the square:
                              $$x^2+ax+a>0 iff left(x+frac a2right)^2-frac{a^2}{4}+a>0 iff \
                              left(x+frac a2right)^2>frac{a^2-4a}{4} iff left(x+frac a2right)^2ge 0>frac{a^2-4a}{4}, 0<a<4.$$







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 25 at 5:03









                              farruhotafarruhota

                              21.2k2841




                              21.2k2841























                                  0












                                  $begingroup$

                                  After completing the square of the above equation you will get:
                                  $$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
                                  which is nothing but:
                                  $$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
                                  The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
                                  Now for the second part
                                  $$a-frac{a^2}{4}gt0$$
                                  which gives $$4a-a^2gt0$$
                                  Or $$a^2-4alt0$$
                                  $$a(a-4)lt0$$ which gives the critical points $a=0,4$
                                  Now this inequality only holds if $0lt alt4$



                                  Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
                                  One more thing is you can check for the vertex of the parabola....
                                  Hope this helps






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    After completing the square of the above equation you will get:
                                    $$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
                                    which is nothing but:
                                    $$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
                                    The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
                                    Now for the second part
                                    $$a-frac{a^2}{4}gt0$$
                                    which gives $$4a-a^2gt0$$
                                    Or $$a^2-4alt0$$
                                    $$a(a-4)lt0$$ which gives the critical points $a=0,4$
                                    Now this inequality only holds if $0lt alt4$



                                    Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
                                    One more thing is you can check for the vertex of the parabola....
                                    Hope this helps






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      After completing the square of the above equation you will get:
                                      $$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
                                      which is nothing but:
                                      $$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
                                      The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
                                      Now for the second part
                                      $$a-frac{a^2}{4}gt0$$
                                      which gives $$4a-a^2gt0$$
                                      Or $$a^2-4alt0$$
                                      $$a(a-4)lt0$$ which gives the critical points $a=0,4$
                                      Now this inequality only holds if $0lt alt4$



                                      Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
                                      One more thing is you can check for the vertex of the parabola....
                                      Hope this helps






                                      share|cite|improve this answer









                                      $endgroup$



                                      After completing the square of the above equation you will get:
                                      $$x^2+ax+a=x^2+2(a/2)(x)+a+(a/2)^2 -(a/2)^2$$
                                      which is nothing but:
                                      $$(x+frac{a}{2})^2 + (a-frac{a^2}{4})$$
                                      The first term is always positive no matter what the value of $x$ and $a$ is inside because it is a perfect square. So $$(x+frac{a}{2})^2gt0$$
                                      Now for the second part
                                      $$a-frac{a^2}{4}gt0$$
                                      which gives $$4a-a^2gt0$$
                                      Or $$a^2-4alt0$$
                                      $$a(a-4)lt0$$ which gives the critical points $a=0,4$
                                      Now this inequality only holds if $0lt alt4$



                                      Geometrically, when you plot this curve it should not intersect the x axis which will lead you to analysing the discriminant of the equation which already has been mentioned in other answers.
                                      One more thing is you can check for the vertex of the parabola....
                                      Hope this helps







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 25 at 5:08









                                      SNEHIL SANYALSNEHIL SANYAL

                                      644110




                                      644110















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