I have the following equation to solve analytically: $ 0.5q+sqrt q-2=0$












0












$begingroup$


I sincerely apologize in advance for the simple question I am asking but I really cannot live with this doubt in my mind.



I have the following equation to solve analytically:
$$ 0.5q+sqrt q -2=0$$



Obviously there are different ways to solve it. If I use the substitution $x=sqrt q$, I can get the solutions to a quadratic-form equation and pick only the positive ones. Then, using the initial definition of $x$, I get the unique solution for $q=1.5279$. However, I also saw a solution which goes as follow:



$$sqrt q =2-0.5q to q=(2-0.5q)^2 to 4+0.25q^2-3q=0,$$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.



However, by plotting both the initial equation and the last one, I obviously notice that we are dealing with different functions.



Can someone please carefully explain why this happens and if one can actually performs the transformation shown in the last line to find the zeros of the initial function?






algebra-precalculus roots







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$endgroup$












  • $begingroup$
    Remember that squaring always introduces extra solutions.
    $endgroup$
    – Claude Leibovici
    Jan 17 at 6:04
















0












$begingroup$


I sincerely apologize in advance for the simple question I am asking but I really cannot live with this doubt in my mind.



I have the following equation to solve analytically:
$$ 0.5q+sqrt q -2=0$$



Obviously there are different ways to solve it. If I use the substitution $x=sqrt q$, I can get the solutions to a quadratic-form equation and pick only the positive ones. Then, using the initial definition of $x$, I get the unique solution for $q=1.5279$. However, I also saw a solution which goes as follow:



$$sqrt q =2-0.5q to q=(2-0.5q)^2 to 4+0.25q^2-3q=0,$$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.



However, by plotting both the initial equation and the last one, I obviously notice that we are dealing with different functions.



Can someone please carefully explain why this happens and if one can actually performs the transformation shown in the last line to find the zeros of the initial function?






algebra-precalculus roots







share|cite|improve this question











$endgroup$












  • $begingroup$
    Remember that squaring always introduces extra solutions.
    $endgroup$
    – Claude Leibovici
    Jan 17 at 6:04














0












0








0





$begingroup$


I sincerely apologize in advance for the simple question I am asking but I really cannot live with this doubt in my mind.



I have the following equation to solve analytically:
$$ 0.5q+sqrt q -2=0$$



Obviously there are different ways to solve it. If I use the substitution $x=sqrt q$, I can get the solutions to a quadratic-form equation and pick only the positive ones. Then, using the initial definition of $x$, I get the unique solution for $q=1.5279$. However, I also saw a solution which goes as follow:



$$sqrt q =2-0.5q to q=(2-0.5q)^2 to 4+0.25q^2-3q=0,$$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.



However, by plotting both the initial equation and the last one, I obviously notice that we are dealing with different functions.



Can someone please carefully explain why this happens and if one can actually performs the transformation shown in the last line to find the zeros of the initial function?






algebra-precalculus roots







share|cite|improve this question











$endgroup$




I sincerely apologize in advance for the simple question I am asking but I really cannot live with this doubt in my mind.



I have the following equation to solve analytically:
$$ 0.5q+sqrt q -2=0$$



Obviously there are different ways to solve it. If I use the substitution $x=sqrt q$, I can get the solutions to a quadratic-form equation and pick only the positive ones. Then, using the initial definition of $x$, I get the unique solution for $q=1.5279$. However, I also saw a solution which goes as follow:



$$sqrt q =2-0.5q to q=(2-0.5q)^2 to 4+0.25q^2-3q=0,$$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.



However, by plotting both the initial equation and the last one, I obviously notice that we are dealing with different functions.



Can someone please carefully explain why this happens and if one can actually performs the transformation shown in the last line to find the zeros of the initial function?






algebra-precalculus roots




algebra-precalculus roots






share|cite|improve this question















share|cite|improve this question













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edited Jan 16 at 19:20









jordan_glen

1




1










asked Jan 16 at 18:43









MatteoMatteo

154




154












  • $begingroup$
    Remember that squaring always introduces extra solutions.
    $endgroup$
    – Claude Leibovici
    Jan 17 at 6:04


















  • $begingroup$
    Remember that squaring always introduces extra solutions.
    $endgroup$
    – Claude Leibovici
    Jan 17 at 6:04
















$begingroup$
Remember that squaring always introduces extra solutions.
$endgroup$
– Claude Leibovici
Jan 17 at 6:04




$begingroup$
Remember that squaring always introduces extra solutions.
$endgroup$
– Claude Leibovici
Jan 17 at 6:04










3 Answers
3






active

oldest

votes


















2












$begingroup$


However, I also saw a solution which goes as follow:



$q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$



which however yields TWO solutions, $q=1.5279$ and $q=10.47$.




Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
$$
0.5 q + q^frac12 - 2
$$
you get a number near $6.4$, so it's definitely not zero.



This is analogous to starting with
$$
q = -2
$$
and squaring both sides to get
$$
q^2 = 4
$$
which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$


        However, I also saw a solution which goes as follow:



        $q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$



        which however yields TWO solutions, $q=1.5279$ and $q=10.47$.




        Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
        $$
        0.5 q + q^frac12 - 2
        $$
        you get a number near $6.4$, so it's definitely not zero.



        This is analogous to starting with
        $$
        q = -2
        $$
        and squaring both sides to get
        $$
        q^2 = 4
        $$
        which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$


          However, I also saw a solution which goes as follow:



          $q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$



          which however yields TWO solutions, $q=1.5279$ and $q=10.47$.




          Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
          $$
          0.5 q + q^frac12 - 2
          $$
          you get a number near $6.4$, so it's definitely not zero.



          This is analogous to starting with
          $$
          q = -2
          $$
          and squaring both sides to get
          $$
          q^2 = 4
          $$
          which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$


            However, I also saw a solution which goes as follow:



            $q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$



            which however yields TWO solutions, $q=1.5279$ and $q=10.47$.




            Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
            $$
            0.5 q + q^frac12 - 2
            $$
            you get a number near $6.4$, so it's definitely not zero.



            This is analogous to starting with
            $$
            q = -2
            $$
            and squaring both sides to get
            $$
            q^2 = 4
            $$
            which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.






            share|cite|improve this answer









            $endgroup$




            However, I also saw a solution which goes as follow:



            $q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$



            which however yields TWO solutions, $q=1.5279$ and $q=10.47$.




            Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
            $$
            0.5 q + q^frac12 - 2
            $$
            you get a number near $6.4$, so it's definitely not zero.



            This is analogous to starting with
            $$
            q = -2
            $$
            and squaring both sides to get
            $$
            q^2 = 4
            $$
            which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 16 at 18:50









            John HughesJohn Hughes

            63.9k24191




            63.9k24191























                0












                $begingroup$

                We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.






                    share|cite|improve this answer









                    $endgroup$



                    We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 16 at 18:46









                    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                    75.1k42865




                    75.1k42865























                        0












                        $begingroup$

                        This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it






                            share|cite|improve this answer









                            $endgroup$



                            This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 16 at 20:12









                            QuaternionQuaternion

                            1




                            1






























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