Why do I keep getting this incorrect solution when trying to find all the real solutions for $sqrt{2x-3}...












4














The problem is to find all real solutions (if any exists) for $sqrt{2x-3} +x=3$.



Now, my textbook says the answer is {2}, however, I keep getting {2, 6}. I've tried multiple approaches, but here is one of them:



I got rid of the root by squaring both sides,
$$sqrt{2x-3}^2=(3-x)^2$$
$$0=12-8x+x^2$$
Using the AC method, I got
$$(-x^2+6x)(2x-12)=0$$
$$-x(x-6)2(x-6)=0$$
$$(-x+2)(x-6)=0$$
hence, $$x=2, x=6$$



Of course, I can always just check my solutions and I'd immediately recognize 6 does not work. But that's a bit too tedious for my taste. Can anyone explain where I went wrong with my approach?










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  • 2




    Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
    – T. Ford
    Dec 29 '18 at 5:52






  • 8




    Possible duplicate of Why do extraneous solutions exist? or When do we get extraneous roots?
    – Carmeister
    Dec 29 '18 at 8:09












  • The simplest way to "debug" such things is to check the solutions at each step, then you can (usually) find out the first wrong step.
    – user202729
    Dec 29 '18 at 9:07








  • 2




    It's not duplicate of course.
    – Michael Rozenberg
    Dec 30 '18 at 19:04
















4














The problem is to find all real solutions (if any exists) for $sqrt{2x-3} +x=3$.



Now, my textbook says the answer is {2}, however, I keep getting {2, 6}. I've tried multiple approaches, but here is one of them:



I got rid of the root by squaring both sides,
$$sqrt{2x-3}^2=(3-x)^2$$
$$0=12-8x+x^2$$
Using the AC method, I got
$$(-x^2+6x)(2x-12)=0$$
$$-x(x-6)2(x-6)=0$$
$$(-x+2)(x-6)=0$$
hence, $$x=2, x=6$$



Of course, I can always just check my solutions and I'd immediately recognize 6 does not work. But that's a bit too tedious for my taste. Can anyone explain where I went wrong with my approach?










share|cite|improve this question




















  • 2




    Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
    – T. Ford
    Dec 29 '18 at 5:52






  • 8




    Possible duplicate of Why do extraneous solutions exist? or When do we get extraneous roots?
    – Carmeister
    Dec 29 '18 at 8:09












  • The simplest way to "debug" such things is to check the solutions at each step, then you can (usually) find out the first wrong step.
    – user202729
    Dec 29 '18 at 9:07








  • 2




    It's not duplicate of course.
    – Michael Rozenberg
    Dec 30 '18 at 19:04














4












4








4







The problem is to find all real solutions (if any exists) for $sqrt{2x-3} +x=3$.



Now, my textbook says the answer is {2}, however, I keep getting {2, 6}. I've tried multiple approaches, but here is one of them:



I got rid of the root by squaring both sides,
$$sqrt{2x-3}^2=(3-x)^2$$
$$0=12-8x+x^2$$
Using the AC method, I got
$$(-x^2+6x)(2x-12)=0$$
$$-x(x-6)2(x-6)=0$$
$$(-x+2)(x-6)=0$$
hence, $$x=2, x=6$$



Of course, I can always just check my solutions and I'd immediately recognize 6 does not work. But that's a bit too tedious for my taste. Can anyone explain where I went wrong with my approach?










share|cite|improve this question















The problem is to find all real solutions (if any exists) for $sqrt{2x-3} +x=3$.



Now, my textbook says the answer is {2}, however, I keep getting {2, 6}. I've tried multiple approaches, but here is one of them:



I got rid of the root by squaring both sides,
$$sqrt{2x-3}^2=(3-x)^2$$
$$0=12-8x+x^2$$
Using the AC method, I got
$$(-x^2+6x)(2x-12)=0$$
$$-x(x-6)2(x-6)=0$$
$$(-x+2)(x-6)=0$$
hence, $$x=2, x=6$$



Of course, I can always just check my solutions and I'd immediately recognize 6 does not work. But that's a bit too tedious for my taste. Can anyone explain where I went wrong with my approach?







algebra-precalculus






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edited Dec 29 '18 at 12:01









Asaf Karagila

302k32426756




302k32426756










asked Dec 29 '18 at 5:45









Lex_i

707




707








  • 2




    Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
    – T. Ford
    Dec 29 '18 at 5:52






  • 8




    Possible duplicate of Why do extraneous solutions exist? or When do we get extraneous roots?
    – Carmeister
    Dec 29 '18 at 8:09












  • The simplest way to "debug" such things is to check the solutions at each step, then you can (usually) find out the first wrong step.
    – user202729
    Dec 29 '18 at 9:07








  • 2




    It's not duplicate of course.
    – Michael Rozenberg
    Dec 30 '18 at 19:04














  • 2




    Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
    – T. Ford
    Dec 29 '18 at 5:52






  • 8




    Possible duplicate of Why do extraneous solutions exist? or When do we get extraneous roots?
    – Carmeister
    Dec 29 '18 at 8:09












  • The simplest way to "debug" such things is to check the solutions at each step, then you can (usually) find out the first wrong step.
    – user202729
    Dec 29 '18 at 9:07








  • 2




    It's not duplicate of course.
    – Michael Rozenberg
    Dec 30 '18 at 19:04








2




2




Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
– T. Ford
Dec 29 '18 at 5:52




Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
– T. Ford
Dec 29 '18 at 5:52




8




8




Possible duplicate of Why do extraneous solutions exist? or When do we get extraneous roots?
– Carmeister
Dec 29 '18 at 8:09






Possible duplicate of Why do extraneous solutions exist? or When do we get extraneous roots?
– Carmeister
Dec 29 '18 at 8:09














The simplest way to "debug" such things is to check the solutions at each step, then you can (usually) find out the first wrong step.
– user202729
Dec 29 '18 at 9:07






The simplest way to "debug" such things is to check the solutions at each step, then you can (usually) find out the first wrong step.
– user202729
Dec 29 '18 at 9:07






2




2




It's not duplicate of course.
– Michael Rozenberg
Dec 30 '18 at 19:04




It's not duplicate of course.
– Michael Rozenberg
Dec 30 '18 at 19:04










6 Answers
6






active

oldest

votes


















16














Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.



As a very simple example, notice the following two equations:



$$x = sqrt 4 iff x = +2$$



$$x^2 = 4 iff vert xvert = 2 iff x = pm 2$$



The first equation has only one solution: $+sqrt 4$. The second, however, has two solutions: $pmsqrt 4$. And you get the second equation by squaring the first one.



The exact same idea applies to your example. You have



$$sqrt{2x-3} = 3-x$$



which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving



$$0 = 12-8x+x^2 iff color{blue}{pm}sqrt{2x-3} = 3-x$$



which has a $pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:



$$2x-3 = 9-6x+x^2; quad color{blue}{x leq 3}$$



$$0 = 12-8x+x^2; quad color{blue}{x leq 3}$$



Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.






share|cite|improve this answer























  • Maybe it would be a little bit more correct to say that the RHS should be non-negative, since the LHS must be non-negative. There is no way to make the LHS negative.
    – Kai
    Dec 29 '18 at 22:47



















10














When we square both sides, we could have introduce additional solution.



An extreme example is as follows:



Solve $x=1$.



The solution is just $x=1$.



However, if we square them, $x^2=1$. Now $x=-1$ also satisfies the new equation which is no longer the original problem.



Remark: Note that as we write $$sqrt{2x-3}=3-x,$$



there is an implicit constraint that we need $3-x ge 0$.






share|cite|improve this answer





























    2














    The initial question is actually:



    If $x$ exists, then it satisfies $sqrt{2x-3}+x=3$. What is $x$?



    With each logically sound algebraic step, the initial question is rephrased, eventually leading to:



    If $x$ exists, then it satisfies $x = 2text{ or } x= 6$. What is $x$?



    Unfortunately, we still have done nothing to prove x exists. If all the logical steps are if and only if , or reversible, then we are done. We could 'let $x = 2$ or $x = 6$' and follow the logic backwards to demonstrate that x is a solution to the original equation. Unfortunately, as noted on other answers, squaring is not a reversible step; the square root function is not the same as the inverse of the square function. We can see this by noting that the square function takes positive and negative numbers and maps them to positive numbers. Meanwhile, the square root function takes positive numbers only and maps them to positive numbers only.



    All this is a long way of saying that the alternative to checking answers is understanding which algebraic steps are reversible and which are not. In practice, its easier to just check your answers every time.






    share|cite|improve this answer































      1














      To build off of the other answers provided, the equation you wish to solve is



      $$sqrt{2x-3} = 3-x$$



      To do so, you square both sides, and solve



      $$2x-3 = (3-x)^2$$



      which has two solutions. However this equation can also be obtained by squaring



      $$-sqrt{2x-3} = 3-x$$



      The second solution is the solution to this second equation. This is easy to see with a plot:



      enter image description here






      share|cite|improve this answer































        1














        Your reasoning is a chain of implications: if $x_0$ is a solution, then..., then $x_0$ should be $cdot$ or $cdot$. As you do not have equivalences instead of implications at each step, the final potential $x$s are only a superset of solutions, that should be plugged into the original problem, to see if they fit.



        A change in variables can show some additional insights. To get rid of the root, you can choose a positive $y$ such that $y^2 = 2x-3$, and thus $ 2x-3=sqrt{y}$.



        You can rewrite your equation as:



        $$y+(y^2+3)/2 = 3$$



        or



        $$2y+y^2 = 3$$



        This system has at most two solutions $y_a$ and $y_b$: an obvious solution is real, satisfying $2times 1+1^2=3$, or $y_a=1$, which yields $x=2$. Since $y_atimes y_b = -3$, the second solution would be negative, which is ruled out by hypothesis.



        So $x=2$ is the only solution.






        share|cite|improve this answer





























          0














          Squaring both sides of an equation can introduce extraneous solutions. For instance, $x=-2$ isn't a solution of $x=2$; but it is a solution of $x^2=4$.



          Thus it is necessary when doing so to check your answer.



          Notice:$$sqrt{2cdot 6-3}+6=9neq3$$.






          share|cite|improve this answer























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            6 Answers
            6






            active

            oldest

            votes








            6 Answers
            6






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            16














            Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.



            As a very simple example, notice the following two equations:



            $$x = sqrt 4 iff x = +2$$



            $$x^2 = 4 iff vert xvert = 2 iff x = pm 2$$



            The first equation has only one solution: $+sqrt 4$. The second, however, has two solutions: $pmsqrt 4$. And you get the second equation by squaring the first one.



            The exact same idea applies to your example. You have



            $$sqrt{2x-3} = 3-x$$



            which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving



            $$0 = 12-8x+x^2 iff color{blue}{pm}sqrt{2x-3} = 3-x$$



            which has a $pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:



            $$2x-3 = 9-6x+x^2; quad color{blue}{x leq 3}$$



            $$0 = 12-8x+x^2; quad color{blue}{x leq 3}$$



            Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.






            share|cite|improve this answer























            • Maybe it would be a little bit more correct to say that the RHS should be non-negative, since the LHS must be non-negative. There is no way to make the LHS negative.
              – Kai
              Dec 29 '18 at 22:47
















            16














            Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.



            As a very simple example, notice the following two equations:



            $$x = sqrt 4 iff x = +2$$



            $$x^2 = 4 iff vert xvert = 2 iff x = pm 2$$



            The first equation has only one solution: $+sqrt 4$. The second, however, has two solutions: $pmsqrt 4$. And you get the second equation by squaring the first one.



            The exact same idea applies to your example. You have



            $$sqrt{2x-3} = 3-x$$



            which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving



            $$0 = 12-8x+x^2 iff color{blue}{pm}sqrt{2x-3} = 3-x$$



            which has a $pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:



            $$2x-3 = 9-6x+x^2; quad color{blue}{x leq 3}$$



            $$0 = 12-8x+x^2; quad color{blue}{x leq 3}$$



            Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.






            share|cite|improve this answer























            • Maybe it would be a little bit more correct to say that the RHS should be non-negative, since the LHS must be non-negative. There is no way to make the LHS negative.
              – Kai
              Dec 29 '18 at 22:47














            16












            16








            16






            Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.



            As a very simple example, notice the following two equations:



            $$x = sqrt 4 iff x = +2$$



            $$x^2 = 4 iff vert xvert = 2 iff x = pm 2$$



            The first equation has only one solution: $+sqrt 4$. The second, however, has two solutions: $pmsqrt 4$. And you get the second equation by squaring the first one.



            The exact same idea applies to your example. You have



            $$sqrt{2x-3} = 3-x$$



            which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving



            $$0 = 12-8x+x^2 iff color{blue}{pm}sqrt{2x-3} = 3-x$$



            which has a $pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:



            $$2x-3 = 9-6x+x^2; quad color{blue}{x leq 3}$$



            $$0 = 12-8x+x^2; quad color{blue}{x leq 3}$$



            Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.






            share|cite|improve this answer














            Because squaring both sides of an equation always introduces the “risk” of an extraneous solution.



            As a very simple example, notice the following two equations:



            $$x = sqrt 4 iff x = +2$$



            $$x^2 = 4 iff vert xvert = 2 iff x = pm 2$$



            The first equation has only one solution: $+sqrt 4$. The second, however, has two solutions: $pmsqrt 4$. And you get the second equation by squaring the first one.



            The exact same idea applies to your example. You have



            $$sqrt{2x-3} = 3-x$$



            which refers only to the non-negative square root of $2x-3$. So, if a solution makes the LHS negative, it is extraneous. But, when you square both sides, you’re actually solving



            $$0 = 12-8x+x^2 iff color{blue}{pm}sqrt{2x-3} = 3-x$$



            which has a $pm$ sign and is therefore not the same equation. Now, to be precise, you’d have to add the condition that the LHS must be non-negative:



            $$2x-3 = 9-6x+x^2; quad color{blue}{x leq 3}$$



            $$0 = 12-8x+x^2; quad color{blue}{x leq 3}$$



            Now, your equation is equivalent to the first with the given constraint. If you get any solution greater than $3$, (in this case, $6$), you’d know it satisfies the new equation but not the original one.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 29 '18 at 6:15

























            answered Dec 29 '18 at 6:02









            KM101

            5,2191423




            5,2191423












            • Maybe it would be a little bit more correct to say that the RHS should be non-negative, since the LHS must be non-negative. There is no way to make the LHS negative.
              – Kai
              Dec 29 '18 at 22:47


















            • Maybe it would be a little bit more correct to say that the RHS should be non-negative, since the LHS must be non-negative. There is no way to make the LHS negative.
              – Kai
              Dec 29 '18 at 22:47
















            Maybe it would be a little bit more correct to say that the RHS should be non-negative, since the LHS must be non-negative. There is no way to make the LHS negative.
            – Kai
            Dec 29 '18 at 22:47




            Maybe it would be a little bit more correct to say that the RHS should be non-negative, since the LHS must be non-negative. There is no way to make the LHS negative.
            – Kai
            Dec 29 '18 at 22:47











            10














            When we square both sides, we could have introduce additional solution.



            An extreme example is as follows:



            Solve $x=1$.



            The solution is just $x=1$.



            However, if we square them, $x^2=1$. Now $x=-1$ also satisfies the new equation which is no longer the original problem.



            Remark: Note that as we write $$sqrt{2x-3}=3-x,$$



            there is an implicit constraint that we need $3-x ge 0$.






            share|cite|improve this answer


























              10














              When we square both sides, we could have introduce additional solution.



              An extreme example is as follows:



              Solve $x=1$.



              The solution is just $x=1$.



              However, if we square them, $x^2=1$. Now $x=-1$ also satisfies the new equation which is no longer the original problem.



              Remark: Note that as we write $$sqrt{2x-3}=3-x,$$



              there is an implicit constraint that we need $3-x ge 0$.






              share|cite|improve this answer
























                10












                10








                10






                When we square both sides, we could have introduce additional solution.



                An extreme example is as follows:



                Solve $x=1$.



                The solution is just $x=1$.



                However, if we square them, $x^2=1$. Now $x=-1$ also satisfies the new equation which is no longer the original problem.



                Remark: Note that as we write $$sqrt{2x-3}=3-x,$$



                there is an implicit constraint that we need $3-x ge 0$.






                share|cite|improve this answer












                When we square both sides, we could have introduce additional solution.



                An extreme example is as follows:



                Solve $x=1$.



                The solution is just $x=1$.



                However, if we square them, $x^2=1$. Now $x=-1$ also satisfies the new equation which is no longer the original problem.



                Remark: Note that as we write $$sqrt{2x-3}=3-x,$$



                there is an implicit constraint that we need $3-x ge 0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 29 '18 at 5:49









                Siong Thye Goh

                99.5k1465117




                99.5k1465117























                    2














                    The initial question is actually:



                    If $x$ exists, then it satisfies $sqrt{2x-3}+x=3$. What is $x$?



                    With each logically sound algebraic step, the initial question is rephrased, eventually leading to:



                    If $x$ exists, then it satisfies $x = 2text{ or } x= 6$. What is $x$?



                    Unfortunately, we still have done nothing to prove x exists. If all the logical steps are if and only if , or reversible, then we are done. We could 'let $x = 2$ or $x = 6$' and follow the logic backwards to demonstrate that x is a solution to the original equation. Unfortunately, as noted on other answers, squaring is not a reversible step; the square root function is not the same as the inverse of the square function. We can see this by noting that the square function takes positive and negative numbers and maps them to positive numbers. Meanwhile, the square root function takes positive numbers only and maps them to positive numbers only.



                    All this is a long way of saying that the alternative to checking answers is understanding which algebraic steps are reversible and which are not. In practice, its easier to just check your answers every time.






                    share|cite|improve this answer




























                      2














                      The initial question is actually:



                      If $x$ exists, then it satisfies $sqrt{2x-3}+x=3$. What is $x$?



                      With each logically sound algebraic step, the initial question is rephrased, eventually leading to:



                      If $x$ exists, then it satisfies $x = 2text{ or } x= 6$. What is $x$?



                      Unfortunately, we still have done nothing to prove x exists. If all the logical steps are if and only if , or reversible, then we are done. We could 'let $x = 2$ or $x = 6$' and follow the logic backwards to demonstrate that x is a solution to the original equation. Unfortunately, as noted on other answers, squaring is not a reversible step; the square root function is not the same as the inverse of the square function. We can see this by noting that the square function takes positive and negative numbers and maps them to positive numbers. Meanwhile, the square root function takes positive numbers only and maps them to positive numbers only.



                      All this is a long way of saying that the alternative to checking answers is understanding which algebraic steps are reversible and which are not. In practice, its easier to just check your answers every time.






                      share|cite|improve this answer


























                        2












                        2








                        2






                        The initial question is actually:



                        If $x$ exists, then it satisfies $sqrt{2x-3}+x=3$. What is $x$?



                        With each logically sound algebraic step, the initial question is rephrased, eventually leading to:



                        If $x$ exists, then it satisfies $x = 2text{ or } x= 6$. What is $x$?



                        Unfortunately, we still have done nothing to prove x exists. If all the logical steps are if and only if , or reversible, then we are done. We could 'let $x = 2$ or $x = 6$' and follow the logic backwards to demonstrate that x is a solution to the original equation. Unfortunately, as noted on other answers, squaring is not a reversible step; the square root function is not the same as the inverse of the square function. We can see this by noting that the square function takes positive and negative numbers and maps them to positive numbers. Meanwhile, the square root function takes positive numbers only and maps them to positive numbers only.



                        All this is a long way of saying that the alternative to checking answers is understanding which algebraic steps are reversible and which are not. In practice, its easier to just check your answers every time.






                        share|cite|improve this answer














                        The initial question is actually:



                        If $x$ exists, then it satisfies $sqrt{2x-3}+x=3$. What is $x$?



                        With each logically sound algebraic step, the initial question is rephrased, eventually leading to:



                        If $x$ exists, then it satisfies $x = 2text{ or } x= 6$. What is $x$?



                        Unfortunately, we still have done nothing to prove x exists. If all the logical steps are if and only if , or reversible, then we are done. We could 'let $x = 2$ or $x = 6$' and follow the logic backwards to demonstrate that x is a solution to the original equation. Unfortunately, as noted on other answers, squaring is not a reversible step; the square root function is not the same as the inverse of the square function. We can see this by noting that the square function takes positive and negative numbers and maps them to positive numbers. Meanwhile, the square root function takes positive numbers only and maps them to positive numbers only.



                        All this is a long way of saying that the alternative to checking answers is understanding which algebraic steps are reversible and which are not. In practice, its easier to just check your answers every time.







                        share|cite|improve this answer














                        share|cite|improve this answer



                        share|cite|improve this answer








                        edited Dec 29 '18 at 22:34

























                        answered Dec 29 '18 at 6:23









                        David Diaz

                        954420




                        954420























                            1














                            To build off of the other answers provided, the equation you wish to solve is



                            $$sqrt{2x-3} = 3-x$$



                            To do so, you square both sides, and solve



                            $$2x-3 = (3-x)^2$$



                            which has two solutions. However this equation can also be obtained by squaring



                            $$-sqrt{2x-3} = 3-x$$



                            The second solution is the solution to this second equation. This is easy to see with a plot:



                            enter image description here






                            share|cite|improve this answer




























                              1














                              To build off of the other answers provided, the equation you wish to solve is



                              $$sqrt{2x-3} = 3-x$$



                              To do so, you square both sides, and solve



                              $$2x-3 = (3-x)^2$$



                              which has two solutions. However this equation can also be obtained by squaring



                              $$-sqrt{2x-3} = 3-x$$



                              The second solution is the solution to this second equation. This is easy to see with a plot:



                              enter image description here






                              share|cite|improve this answer


























                                1












                                1








                                1






                                To build off of the other answers provided, the equation you wish to solve is



                                $$sqrt{2x-3} = 3-x$$



                                To do so, you square both sides, and solve



                                $$2x-3 = (3-x)^2$$



                                which has two solutions. However this equation can also be obtained by squaring



                                $$-sqrt{2x-3} = 3-x$$



                                The second solution is the solution to this second equation. This is easy to see with a plot:



                                enter image description here






                                share|cite|improve this answer














                                To build off of the other answers provided, the equation you wish to solve is



                                $$sqrt{2x-3} = 3-x$$



                                To do so, you square both sides, and solve



                                $$2x-3 = (3-x)^2$$



                                which has two solutions. However this equation can also be obtained by squaring



                                $$-sqrt{2x-3} = 3-x$$



                                The second solution is the solution to this second equation. This is easy to see with a plot:



                                enter image description here







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Dec 29 '18 at 8:32

























                                answered Dec 29 '18 at 8:21









                                Kai

                                22326




                                22326























                                    1














                                    Your reasoning is a chain of implications: if $x_0$ is a solution, then..., then $x_0$ should be $cdot$ or $cdot$. As you do not have equivalences instead of implications at each step, the final potential $x$s are only a superset of solutions, that should be plugged into the original problem, to see if they fit.



                                    A change in variables can show some additional insights. To get rid of the root, you can choose a positive $y$ such that $y^2 = 2x-3$, and thus $ 2x-3=sqrt{y}$.



                                    You can rewrite your equation as:



                                    $$y+(y^2+3)/2 = 3$$



                                    or



                                    $$2y+y^2 = 3$$



                                    This system has at most two solutions $y_a$ and $y_b$: an obvious solution is real, satisfying $2times 1+1^2=3$, or $y_a=1$, which yields $x=2$. Since $y_atimes y_b = -3$, the second solution would be negative, which is ruled out by hypothesis.



                                    So $x=2$ is the only solution.






                                    share|cite|improve this answer


























                                      1














                                      Your reasoning is a chain of implications: if $x_0$ is a solution, then..., then $x_0$ should be $cdot$ or $cdot$. As you do not have equivalences instead of implications at each step, the final potential $x$s are only a superset of solutions, that should be plugged into the original problem, to see if they fit.



                                      A change in variables can show some additional insights. To get rid of the root, you can choose a positive $y$ such that $y^2 = 2x-3$, and thus $ 2x-3=sqrt{y}$.



                                      You can rewrite your equation as:



                                      $$y+(y^2+3)/2 = 3$$



                                      or



                                      $$2y+y^2 = 3$$



                                      This system has at most two solutions $y_a$ and $y_b$: an obvious solution is real, satisfying $2times 1+1^2=3$, or $y_a=1$, which yields $x=2$. Since $y_atimes y_b = -3$, the second solution would be negative, which is ruled out by hypothesis.



                                      So $x=2$ is the only solution.






                                      share|cite|improve this answer
























                                        1












                                        1








                                        1






                                        Your reasoning is a chain of implications: if $x_0$ is a solution, then..., then $x_0$ should be $cdot$ or $cdot$. As you do not have equivalences instead of implications at each step, the final potential $x$s are only a superset of solutions, that should be plugged into the original problem, to see if they fit.



                                        A change in variables can show some additional insights. To get rid of the root, you can choose a positive $y$ such that $y^2 = 2x-3$, and thus $ 2x-3=sqrt{y}$.



                                        You can rewrite your equation as:



                                        $$y+(y^2+3)/2 = 3$$



                                        or



                                        $$2y+y^2 = 3$$



                                        This system has at most two solutions $y_a$ and $y_b$: an obvious solution is real, satisfying $2times 1+1^2=3$, or $y_a=1$, which yields $x=2$. Since $y_atimes y_b = -3$, the second solution would be negative, which is ruled out by hypothesis.



                                        So $x=2$ is the only solution.






                                        share|cite|improve this answer












                                        Your reasoning is a chain of implications: if $x_0$ is a solution, then..., then $x_0$ should be $cdot$ or $cdot$. As you do not have equivalences instead of implications at each step, the final potential $x$s are only a superset of solutions, that should be plugged into the original problem, to see if they fit.



                                        A change in variables can show some additional insights. To get rid of the root, you can choose a positive $y$ such that $y^2 = 2x-3$, and thus $ 2x-3=sqrt{y}$.



                                        You can rewrite your equation as:



                                        $$y+(y^2+3)/2 = 3$$



                                        or



                                        $$2y+y^2 = 3$$



                                        This system has at most two solutions $y_a$ and $y_b$: an obvious solution is real, satisfying $2times 1+1^2=3$, or $y_a=1$, which yields $x=2$. Since $y_atimes y_b = -3$, the second solution would be negative, which is ruled out by hypothesis.



                                        So $x=2$ is the only solution.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered Dec 30 '18 at 21:48









                                        Laurent Duval

                                        5,29811239




                                        5,29811239























                                            0














                                            Squaring both sides of an equation can introduce extraneous solutions. For instance, $x=-2$ isn't a solution of $x=2$; but it is a solution of $x^2=4$.



                                            Thus it is necessary when doing so to check your answer.



                                            Notice:$$sqrt{2cdot 6-3}+6=9neq3$$.






                                            share|cite|improve this answer




























                                              0














                                              Squaring both sides of an equation can introduce extraneous solutions. For instance, $x=-2$ isn't a solution of $x=2$; but it is a solution of $x^2=4$.



                                              Thus it is necessary when doing so to check your answer.



                                              Notice:$$sqrt{2cdot 6-3}+6=9neq3$$.






                                              share|cite|improve this answer


























                                                0












                                                0








                                                0






                                                Squaring both sides of an equation can introduce extraneous solutions. For instance, $x=-2$ isn't a solution of $x=2$; but it is a solution of $x^2=4$.



                                                Thus it is necessary when doing so to check your answer.



                                                Notice:$$sqrt{2cdot 6-3}+6=9neq3$$.






                                                share|cite|improve this answer














                                                Squaring both sides of an equation can introduce extraneous solutions. For instance, $x=-2$ isn't a solution of $x=2$; but it is a solution of $x^2=4$.



                                                Thus it is necessary when doing so to check your answer.



                                                Notice:$$sqrt{2cdot 6-3}+6=9neq3$$.







                                                share|cite|improve this answer














                                                share|cite|improve this answer



                                                share|cite|improve this answer








                                                edited yesterday

























                                                answered Dec 29 '18 at 5:57









                                                Chris Custer

                                                10.9k3824




                                                10.9k3824






























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