When I complete the square on $3x^2 - 12x + 14$ I get an imaginary number, where have I gone wrong?












0












$begingroup$


I have a question in my excersise book:




By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$




My approach was to complete the square and rearrange to make $x$ the subject.



The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.



However I get a negative square root:



$$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
$$(sqrt {3}x - 2sqrt{3})^2 = -2$$
$$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
$$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
$$x = (2sqrt{3} +- sqrt {-2})/3$$



Bad formatting: $+-$ means either $+$ or $-$



Where have I gone wrong?






quadratics square-numbers cubic-equations







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    0












    $begingroup$


    I have a question in my excersise book:




    By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$




    My approach was to complete the square and rearrange to make $x$ the subject.



    The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.



    However I get a negative square root:



    $$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
    $$(sqrt {3}x - 2sqrt{3})^2 = -2$$
    $$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
    $$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
    $$x = (2sqrt{3} +- sqrt {-2})/3$$



    Bad formatting: $+-$ means either $+$ or $-$



    Where have I gone wrong?






    quadratics square-numbers cubic-equations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have a question in my excersise book:




      By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$




      My approach was to complete the square and rearrange to make $x$ the subject.



      The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.



      However I get a negative square root:



      $$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
      $$(sqrt {3}x - 2sqrt{3})^2 = -2$$
      $$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
      $$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
      $$x = (2sqrt{3} +- sqrt {-2})/3$$



      Bad formatting: $+-$ means either $+$ or $-$



      Where have I gone wrong?






      quadratics square-numbers cubic-equations







      share|cite|improve this question









      $endgroup$




      I have a question in my excersise book:




      By completing the square show that the expression $3x^2 - 12x + 14$ is positive for all $x$




      My approach was to complete the square and rearrange to make $x$ the subject.



      The answer I came to after completing the square was $(sqrt {3}x - 2sqrt{3})^2+2$.



      However I get a negative square root:



      $$(sqrt {3}x - 2sqrt{3})^2+2 = 0$$
      $$(sqrt {3}x - 2sqrt{3})^2 = -2$$
      $$sqrt {3}x - 2sqrt{3} = sqrt{-2}$$
      $$sqrt{3}x = 2sqrt{3} +- sqrt {-2}$$
      $$x = (2sqrt{3} +- sqrt {-2})/3$$



      Bad formatting: $+-$ means either $+$ or $-$



      Where have I gone wrong?






      quadratics square-numbers cubic-equations




      quadratics square-numbers cubic-equations






      share|cite|improve this question













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      asked Dec 3 '18 at 12:10









      SimonSimon

      1033




      1033






















          4 Answers
          4






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          3












          $begingroup$

          You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
          $$
          (sqrt 3x - 2sqrt3)^2 + 2
          $$
          then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
            $endgroup$
            – Simon
            Dec 3 '18 at 12:21



















          3












          $begingroup$

          You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.






            share|cite|improve this answer









            $endgroup$





















              1












              $begingroup$

              In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.






              share|cite|improve this answer









              $endgroup$













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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                3












                $begingroup$

                You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
                $$
                (sqrt 3x - 2sqrt3)^2 + 2
                $$
                then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                  $endgroup$
                  – Simon
                  Dec 3 '18 at 12:21
















                3












                $begingroup$

                You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
                $$
                (sqrt 3x - 2sqrt3)^2 + 2
                $$
                then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                  $endgroup$
                  – Simon
                  Dec 3 '18 at 12:21














                3












                3








                3





                $begingroup$

                You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
                $$
                (sqrt 3x - 2sqrt3)^2 + 2
                $$
                then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.






                share|cite|improve this answer









                $endgroup$



                You haven't gone wrong per se. You've just gone a step too far. No need to solve the equation or factor anything. Just note that when you have
                $$
                (sqrt 3x - 2sqrt3)^2 + 2
                $$
                then that's a square (which is non-negative) plus $2$, which necessarily makes the value of the entire expression strictly positive, no matter what $x$ is.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 3 '18 at 12:17









                ArthurArthur

                113k7110193




                113k7110193












                • $begingroup$
                  Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                  $endgroup$
                  – Simon
                  Dec 3 '18 at 12:21


















                • $begingroup$
                  Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                  $endgroup$
                  – Simon
                  Dec 3 '18 at 12:21
















                $begingroup$
                Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                $endgroup$
                – Simon
                Dec 3 '18 at 12:21




                $begingroup$
                Thank you. That was much easier than I expected it to be, I'll accept your answer as soon as I can
                $endgroup$
                – Simon
                Dec 3 '18 at 12:21











                3












                $begingroup$

                You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.






                    share|cite|improve this answer









                    $endgroup$



                    You haven't gone wrong. By finding the expression for $x$ after completing the square, you are looking for solutions to the equation $3x^2 - 12x + 14 = 0$. You find that there are no (real) solutions, which means that the graph of this parabola never touches the $x$-axis. Because this is a "valley parabola", certainly it will be positive somewhere; that means it is always positive, because it never crosses or touches the $x$-axis.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 3 '18 at 12:20









                    Mees de VriesMees de Vries

                    16.6k12654




                    16.6k12654























                        2












                        $begingroup$

                        Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.






                            share|cite|improve this answer









                            $endgroup$



                            Do you have to do it exactly that way? Another is to take the factor of 3 (the coefficient of $x^2$) outside, & put it back at the end, to get $$3(x-2)^2+2 .$$ Certainly, this shows just as well that the expression is always positive.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 3 '18 at 12:38









                            AmbretteOrriseyAmbretteOrrisey

                            54210




                            54210























                                1












                                $begingroup$

                                In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.






                                share|cite|improve this answer









                                $endgroup$


















                                  1












                                  $begingroup$

                                  In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.






                                  share|cite|improve this answer









                                  $endgroup$
















                                    1












                                    1








                                    1





                                    $begingroup$

                                    In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.






                                    share|cite|improve this answer









                                    $endgroup$



                                    In general, remember that a negative radicand does not imply you went wrong anywhere. If anything, it is simply another way to point out that the graph never crosses the $x$-axis (hence no real roots), and since $a > 0$, the graph lies entirely above the $x$-axis, which means the quadratic is always positive.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered Dec 3 '18 at 12:25









                                    KM101KM101

                                    5,9481524




                                    5,9481524






























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