How to quickly find an $x$ that works?












0












$begingroup$


When asked to prove if this is true or not $exists x inBbb R$, $x^2 + 29x + 209 leq 0$, it can easily be done by saying when $x = -15$ it works. But how to find that $x = -15$?



For any "there exists" problem that shows up on a test, how to find one that just works?






algebra-precalculus







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    When asked to prove if this is true or not $exists x inBbb R$, $x^2 + 29x + 209 leq 0$, it can easily be done by saying when $x = -15$ it works. But how to find that $x = -15$?



    For any "there exists" problem that shows up on a test, how to find one that just works?






    algebra-precalculus







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      When asked to prove if this is true or not $exists x inBbb R$, $x^2 + 29x + 209 leq 0$, it can easily be done by saying when $x = -15$ it works. But how to find that $x = -15$?



      For any "there exists" problem that shows up on a test, how to find one that just works?






      algebra-precalculus







      share|cite|improve this question











      $endgroup$




      When asked to prove if this is true or not $exists x inBbb R$, $x^2 + 29x + 209 leq 0$, it can easily be done by saying when $x = -15$ it works. But how to find that $x = -15$?



      For any "there exists" problem that shows up on a test, how to find one that just works?






      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 Dec 1 '18 at 6:12









      Saad

      19.7k92352




      19.7k92352










      asked Dec 1 '18 at 5:55









      mingming

      3415




      3415






















          6 Answers
          6






          active

          oldest

          votes


















          1












          $begingroup$

          Solve for the equality and then find the roots, then by dropping the value of the roots will yield the inequality.



          Here $x^2+29x+209=0$ gives, $x=frac{-29pm sqrt{29^2-4times 209}}{2}$ i.e. $x=frac{-29pm sqrt{5}}{2}$. For these values the equality holds.



          Now, taking $sqrt5approx 2$ we get, $xapprox frac{-31}{2}=-15.5$ or, $frac{-27}{2}=-13.5$, then for these values we have a strict inequality. You may take integer part for strict inequality.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I thought this too but we are not allowed calculators in our exams
            $endgroup$
            – ming
            Dec 1 '18 at 6:28










          • $begingroup$
            @rowcol You don't need calculators to find the square root of 5. I didn't as you can see me roughly used the value $sqrt5approx2$
            $endgroup$
            – Sujit Bhattacharyya
            Dec 1 '18 at 12:45



















          1












          $begingroup$

          Note that $f(x)=x^2+29x+209$ is a quadric, then you could compute its roots $r_{1}=-dfrac{29-sqrt{5}}{2}$ and $r_{2}=-dfrac{29+sqrt{5}}{2}$. Then you can verify that $f(x) leq 0$ when $r_{1} leq x leq r_{2}$.



          A way to do this, is factoring and evaluating when $f$ is positive and when $f$ is negative.



          Another way is using Rolle's theorem and the continuity of $f$.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Going back to the original question – proving if $x in mathbb R$, $x^2+29x+209 ≤0$ is true, you don't need to find an $x$ that works.



            The discriminant $b^2-4ac$ is $(29)^2-4(1)(209) > 0$, so there are two real roots. Since all polynomials are differentiable, there must be some $x$ that is $≤ 0$ between the two roots.






            share|cite|improve this answer











            $endgroup$





















              1












              $begingroup$

              The vertex of a parabola $ax^2+bx+c$ is at $-frac b{2a}$. Your parabola opens upward because $a gt 0$, so if any point has a value less than zero the vertex does. You can just check whether $-14.5$ leads to a value less than zero. If it does, you have found a value. If not, there is none.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                But we aren't allowed calculators in our exams and this seems a little tough
                $endgroup$
                – ming
                Dec 1 '18 at 6:28










              • $begingroup$
                @ming: see the new answer.
                $endgroup$
                – Ross Millikan
                Dec 1 '18 at 14:23



















              0












              $begingroup$

              $y=x^2+29x +209$, a parabola , opening upward, has a minimum at the vertex.



              Complete the square:



              $y = (x+29/2)^2-(29/2)^2 +209.$



              Choose $x=-29/2$ (why?).



              P.S. Check if $-(29/2)^2+209 le 0.$






              share|cite|improve this answer









              $endgroup$





















                0












                $begingroup$

                If it were me, I'd say the numbers are a bit large and try to substitute to fix that up. Since $209$ is between $14^2$ and $15^2,$ I might try $x=z-14$ or $x=z-15$. Take the former:



                $$x^2+29x+209 = (z-14)^2+29(z-14)+209$$ $$ =z^2+z-1.$$



                So it's now easy to see $z=0$ gives a negative value. So that means $x=14$ works in the original expression.






                share|cite|improve this answer









                $endgroup$













                  Your Answer





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                  6 Answers
                  6






                  active

                  oldest

                  votes








                  6 Answers
                  6






                  active

                  oldest

                  votes









                  active

                  oldest

                  votes






                  active

                  oldest

                  votes









                  1












                  $begingroup$

                  Solve for the equality and then find the roots, then by dropping the value of the roots will yield the inequality.



                  Here $x^2+29x+209=0$ gives, $x=frac{-29pm sqrt{29^2-4times 209}}{2}$ i.e. $x=frac{-29pm sqrt{5}}{2}$. For these values the equality holds.



                  Now, taking $sqrt5approx 2$ we get, $xapprox frac{-31}{2}=-15.5$ or, $frac{-27}{2}=-13.5$, then for these values we have a strict inequality. You may take integer part for strict inequality.






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    I thought this too but we are not allowed calculators in our exams
                    $endgroup$
                    – ming
                    Dec 1 '18 at 6:28










                  • $begingroup$
                    @rowcol You don't need calculators to find the square root of 5. I didn't as you can see me roughly used the value $sqrt5approx2$
                    $endgroup$
                    – Sujit Bhattacharyya
                    Dec 1 '18 at 12:45
















                  1












                  $begingroup$

                  Solve for the equality and then find the roots, then by dropping the value of the roots will yield the inequality.



                  Here $x^2+29x+209=0$ gives, $x=frac{-29pm sqrt{29^2-4times 209}}{2}$ i.e. $x=frac{-29pm sqrt{5}}{2}$. For these values the equality holds.



                  Now, taking $sqrt5approx 2$ we get, $xapprox frac{-31}{2}=-15.5$ or, $frac{-27}{2}=-13.5$, then for these values we have a strict inequality. You may take integer part for strict inequality.






                  share|cite|improve this answer









                  $endgroup$













                  • $begingroup$
                    I thought this too but we are not allowed calculators in our exams
                    $endgroup$
                    – ming
                    Dec 1 '18 at 6:28










                  • $begingroup$
                    @rowcol You don't need calculators to find the square root of 5. I didn't as you can see me roughly used the value $sqrt5approx2$
                    $endgroup$
                    – Sujit Bhattacharyya
                    Dec 1 '18 at 12:45














                  1












                  1








                  1





                  $begingroup$

                  Solve for the equality and then find the roots, then by dropping the value of the roots will yield the inequality.



                  Here $x^2+29x+209=0$ gives, $x=frac{-29pm sqrt{29^2-4times 209}}{2}$ i.e. $x=frac{-29pm sqrt{5}}{2}$. For these values the equality holds.



                  Now, taking $sqrt5approx 2$ we get, $xapprox frac{-31}{2}=-15.5$ or, $frac{-27}{2}=-13.5$, then for these values we have a strict inequality. You may take integer part for strict inequality.






                  share|cite|improve this answer









                  $endgroup$



                  Solve for the equality and then find the roots, then by dropping the value of the roots will yield the inequality.



                  Here $x^2+29x+209=0$ gives, $x=frac{-29pm sqrt{29^2-4times 209}}{2}$ i.e. $x=frac{-29pm sqrt{5}}{2}$. For these values the equality holds.



                  Now, taking $sqrt5approx 2$ we get, $xapprox frac{-31}{2}=-15.5$ or, $frac{-27}{2}=-13.5$, then for these values we have a strict inequality. You may take integer part for strict inequality.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 1 '18 at 6:06









                  Sujit BhattacharyyaSujit Bhattacharyya

                  1,141419




                  1,141419












                  • $begingroup$
                    I thought this too but we are not allowed calculators in our exams
                    $endgroup$
                    – ming
                    Dec 1 '18 at 6:28










                  • $begingroup$
                    @rowcol You don't need calculators to find the square root of 5. I didn't as you can see me roughly used the value $sqrt5approx2$
                    $endgroup$
                    – Sujit Bhattacharyya
                    Dec 1 '18 at 12:45


















                  • $begingroup$
                    I thought this too but we are not allowed calculators in our exams
                    $endgroup$
                    – ming
                    Dec 1 '18 at 6:28










                  • $begingroup$
                    @rowcol You don't need calculators to find the square root of 5. I didn't as you can see me roughly used the value $sqrt5approx2$
                    $endgroup$
                    – Sujit Bhattacharyya
                    Dec 1 '18 at 12:45
















                  $begingroup$
                  I thought this too but we are not allowed calculators in our exams
                  $endgroup$
                  – ming
                  Dec 1 '18 at 6:28




                  $begingroup$
                  I thought this too but we are not allowed calculators in our exams
                  $endgroup$
                  – ming
                  Dec 1 '18 at 6:28












                  $begingroup$
                  @rowcol You don't need calculators to find the square root of 5. I didn't as you can see me roughly used the value $sqrt5approx2$
                  $endgroup$
                  – Sujit Bhattacharyya
                  Dec 1 '18 at 12:45




                  $begingroup$
                  @rowcol You don't need calculators to find the square root of 5. I didn't as you can see me roughly used the value $sqrt5approx2$
                  $endgroup$
                  – Sujit Bhattacharyya
                  Dec 1 '18 at 12:45











                  1












                  $begingroup$

                  Note that $f(x)=x^2+29x+209$ is a quadric, then you could compute its roots $r_{1}=-dfrac{29-sqrt{5}}{2}$ and $r_{2}=-dfrac{29+sqrt{5}}{2}$. Then you can verify that $f(x) leq 0$ when $r_{1} leq x leq r_{2}$.



                  A way to do this, is factoring and evaluating when $f$ is positive and when $f$ is negative.



                  Another way is using Rolle's theorem and the continuity of $f$.






                  share|cite|improve this answer









                  $endgroup$


















                    1












                    $begingroup$

                    Note that $f(x)=x^2+29x+209$ is a quadric, then you could compute its roots $r_{1}=-dfrac{29-sqrt{5}}{2}$ and $r_{2}=-dfrac{29+sqrt{5}}{2}$. Then you can verify that $f(x) leq 0$ when $r_{1} leq x leq r_{2}$.



                    A way to do this, is factoring and evaluating when $f$ is positive and when $f$ is negative.



                    Another way is using Rolle's theorem and the continuity of $f$.






                    share|cite|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$

                      Note that $f(x)=x^2+29x+209$ is a quadric, then you could compute its roots $r_{1}=-dfrac{29-sqrt{5}}{2}$ and $r_{2}=-dfrac{29+sqrt{5}}{2}$. Then you can verify that $f(x) leq 0$ when $r_{1} leq x leq r_{2}$.



                      A way to do this, is factoring and evaluating when $f$ is positive and when $f$ is negative.



                      Another way is using Rolle's theorem and the continuity of $f$.






                      share|cite|improve this answer









                      $endgroup$



                      Note that $f(x)=x^2+29x+209$ is a quadric, then you could compute its roots $r_{1}=-dfrac{29-sqrt{5}}{2}$ and $r_{2}=-dfrac{29+sqrt{5}}{2}$. Then you can verify that $f(x) leq 0$ when $r_{1} leq x leq r_{2}$.



                      A way to do this, is factoring and evaluating when $f$ is positive and when $f$ is negative.



                      Another way is using Rolle's theorem and the continuity of $f$.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Dec 1 '18 at 6:07









                      rowcolrowcol

                      734




                      734























                          1












                          $begingroup$

                          Going back to the original question – proving if $x in mathbb R$, $x^2+29x+209 ≤0$ is true, you don't need to find an $x$ that works.



                          The discriminant $b^2-4ac$ is $(29)^2-4(1)(209) > 0$, so there are two real roots. Since all polynomials are differentiable, there must be some $x$ that is $≤ 0$ between the two roots.






                          share|cite|improve this answer











                          $endgroup$


















                            1












                            $begingroup$

                            Going back to the original question – proving if $x in mathbb R$, $x^2+29x+209 ≤0$ is true, you don't need to find an $x$ that works.



                            The discriminant $b^2-4ac$ is $(29)^2-4(1)(209) > 0$, so there are two real roots. Since all polynomials are differentiable, there must be some $x$ that is $≤ 0$ between the two roots.






                            share|cite|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              Going back to the original question – proving if $x in mathbb R$, $x^2+29x+209 ≤0$ is true, you don't need to find an $x$ that works.



                              The discriminant $b^2-4ac$ is $(29)^2-4(1)(209) > 0$, so there are two real roots. Since all polynomials are differentiable, there must be some $x$ that is $≤ 0$ between the two roots.






                              share|cite|improve this answer











                              $endgroup$



                              Going back to the original question – proving if $x in mathbb R$, $x^2+29x+209 ≤0$ is true, you don't need to find an $x$ that works.



                              The discriminant $b^2-4ac$ is $(29)^2-4(1)(209) > 0$, so there are two real roots. Since all polynomials are differentiable, there must be some $x$ that is $≤ 0$ between the two roots.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 1 '18 at 7:03

























                              answered Dec 1 '18 at 6:18









                              Toby MakToby Mak

                              3,49811128




                              3,49811128























                                  1












                                  $begingroup$

                                  The vertex of a parabola $ax^2+bx+c$ is at $-frac b{2a}$. Your parabola opens upward because $a gt 0$, so if any point has a value less than zero the vertex does. You can just check whether $-14.5$ leads to a value less than zero. If it does, you have found a value. If not, there is none.






                                  share|cite|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    But we aren't allowed calculators in our exams and this seems a little tough
                                    $endgroup$
                                    – ming
                                    Dec 1 '18 at 6:28










                                  • $begingroup$
                                    @ming: see the new answer.
                                    $endgroup$
                                    – Ross Millikan
                                    Dec 1 '18 at 14:23
















                                  1












                                  $begingroup$

                                  The vertex of a parabola $ax^2+bx+c$ is at $-frac b{2a}$. Your parabola opens upward because $a gt 0$, so if any point has a value less than zero the vertex does. You can just check whether $-14.5$ leads to a value less than zero. If it does, you have found a value. If not, there is none.






                                  share|cite|improve this answer











                                  $endgroup$













                                  • $begingroup$
                                    But we aren't allowed calculators in our exams and this seems a little tough
                                    $endgroup$
                                    – ming
                                    Dec 1 '18 at 6:28










                                  • $begingroup$
                                    @ming: see the new answer.
                                    $endgroup$
                                    – Ross Millikan
                                    Dec 1 '18 at 14:23














                                  1












                                  1








                                  1





                                  $begingroup$

                                  The vertex of a parabola $ax^2+bx+c$ is at $-frac b{2a}$. Your parabola opens upward because $a gt 0$, so if any point has a value less than zero the vertex does. You can just check whether $-14.5$ leads to a value less than zero. If it does, you have found a value. If not, there is none.






                                  share|cite|improve this answer











                                  $endgroup$



                                  The vertex of a parabola $ax^2+bx+c$ is at $-frac b{2a}$. Your parabola opens upward because $a gt 0$, so if any point has a value less than zero the vertex does. You can just check whether $-14.5$ leads to a value less than zero. If it does, you have found a value. If not, there is none.







                                  share|cite|improve this answer














                                  share|cite|improve this answer



                                  share|cite|improve this answer








                                  edited Dec 1 '18 at 14:23

























                                  answered Dec 1 '18 at 6:04









                                  Ross MillikanRoss Millikan

                                  296k23198371




                                  296k23198371












                                  • $begingroup$
                                    But we aren't allowed calculators in our exams and this seems a little tough
                                    $endgroup$
                                    – ming
                                    Dec 1 '18 at 6:28










                                  • $begingroup$
                                    @ming: see the new answer.
                                    $endgroup$
                                    – Ross Millikan
                                    Dec 1 '18 at 14:23


















                                  • $begingroup$
                                    But we aren't allowed calculators in our exams and this seems a little tough
                                    $endgroup$
                                    – ming
                                    Dec 1 '18 at 6:28










                                  • $begingroup$
                                    @ming: see the new answer.
                                    $endgroup$
                                    – Ross Millikan
                                    Dec 1 '18 at 14:23
















                                  $begingroup$
                                  But we aren't allowed calculators in our exams and this seems a little tough
                                  $endgroup$
                                  – ming
                                  Dec 1 '18 at 6:28




                                  $begingroup$
                                  But we aren't allowed calculators in our exams and this seems a little tough
                                  $endgroup$
                                  – ming
                                  Dec 1 '18 at 6:28












                                  $begingroup$
                                  @ming: see the new answer.
                                  $endgroup$
                                  – Ross Millikan
                                  Dec 1 '18 at 14:23




                                  $begingroup$
                                  @ming: see the new answer.
                                  $endgroup$
                                  – Ross Millikan
                                  Dec 1 '18 at 14:23











                                  0












                                  $begingroup$

                                  $y=x^2+29x +209$, a parabola , opening upward, has a minimum at the vertex.



                                  Complete the square:



                                  $y = (x+29/2)^2-(29/2)^2 +209.$



                                  Choose $x=-29/2$ (why?).



                                  P.S. Check if $-(29/2)^2+209 le 0.$






                                  share|cite|improve this answer









                                  $endgroup$


















                                    0












                                    $begingroup$

                                    $y=x^2+29x +209$, a parabola , opening upward, has a minimum at the vertex.



                                    Complete the square:



                                    $y = (x+29/2)^2-(29/2)^2 +209.$



                                    Choose $x=-29/2$ (why?).



                                    P.S. Check if $-(29/2)^2+209 le 0.$






                                    share|cite|improve this answer









                                    $endgroup$
















                                      0












                                      0








                                      0





                                      $begingroup$

                                      $y=x^2+29x +209$, a parabola , opening upward, has a minimum at the vertex.



                                      Complete the square:



                                      $y = (x+29/2)^2-(29/2)^2 +209.$



                                      Choose $x=-29/2$ (why?).



                                      P.S. Check if $-(29/2)^2+209 le 0.$






                                      share|cite|improve this answer









                                      $endgroup$



                                      $y=x^2+29x +209$, a parabola , opening upward, has a minimum at the vertex.



                                      Complete the square:



                                      $y = (x+29/2)^2-(29/2)^2 +209.$



                                      Choose $x=-29/2$ (why?).



                                      P.S. Check if $-(29/2)^2+209 le 0.$







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Dec 1 '18 at 9:00









                                      Peter SzilasPeter Szilas

                                      11.3k2822




                                      11.3k2822























                                          0












                                          $begingroup$

                                          If it were me, I'd say the numbers are a bit large and try to substitute to fix that up. Since $209$ is between $14^2$ and $15^2,$ I might try $x=z-14$ or $x=z-15$. Take the former:



                                          $$x^2+29x+209 = (z-14)^2+29(z-14)+209$$ $$ =z^2+z-1.$$



                                          So it's now easy to see $z=0$ gives a negative value. So that means $x=14$ works in the original expression.






                                          share|cite|improve this answer









                                          $endgroup$


















                                            0












                                            $begingroup$

                                            If it were me, I'd say the numbers are a bit large and try to substitute to fix that up. Since $209$ is between $14^2$ and $15^2,$ I might try $x=z-14$ or $x=z-15$. Take the former:



                                            $$x^2+29x+209 = (z-14)^2+29(z-14)+209$$ $$ =z^2+z-1.$$



                                            So it's now easy to see $z=0$ gives a negative value. So that means $x=14$ works in the original expression.






                                            share|cite|improve this answer









                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              If it were me, I'd say the numbers are a bit large and try to substitute to fix that up. Since $209$ is between $14^2$ and $15^2,$ I might try $x=z-14$ or $x=z-15$. Take the former:



                                              $$x^2+29x+209 = (z-14)^2+29(z-14)+209$$ $$ =z^2+z-1.$$



                                              So it's now easy to see $z=0$ gives a negative value. So that means $x=14$ works in the original expression.






                                              share|cite|improve this answer









                                              $endgroup$



                                              If it were me, I'd say the numbers are a bit large and try to substitute to fix that up. Since $209$ is between $14^2$ and $15^2,$ I might try $x=z-14$ or $x=z-15$. Take the former:



                                              $$x^2+29x+209 = (z-14)^2+29(z-14)+209$$ $$ =z^2+z-1.$$



                                              So it's now easy to see $z=0$ gives a negative value. So that means $x=14$ works in the original expression.







                                              share|cite|improve this answer












                                              share|cite|improve this answer



                                              share|cite|improve this answer










                                              answered Dec 1 '18 at 14:26









                                              B. GoddardB. Goddard

                                              18.9k21440




                                              18.9k21440






























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