Why do I keep getting this incorrect solution for this polynomial problem?












1














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




















  • 1




    Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
    – T. Ford
    1 hour ago
















1














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




















  • 1




    Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
    – T. Ford
    1 hour ago














1












1








1







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 55 mins ago

























asked 1 hour ago









Lex_i

526




526








  • 1




    Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
    – T. Ford
    1 hour ago














  • 1




    Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
    – T. Ford
    1 hour ago








1




1




Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
– T. Ford
1 hour ago




Small typo: $sqrt{2x-2}^2$ should read $sqrt{2x-3}^2$.
– T. Ford
1 hour ago










5 Answers
5






active

oldest

votes


















6














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































    8














    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





























      1














      Squaring both sides of an equation can introduce extraneous solutions. Thus it is necessary when doing so to check your answer.



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






      share|cite|improve this answer





























        0














        The initial question is actually:



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



        With each algebraic step, if-then logic is used to rephrase the initial question, eventually leading to:



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



        If all the logical steps are 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 inverse of the square root function is not the same as 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, this is much harder than just checking every answer.






        share|cite|improve this answer





























          0














          We can use also the following way.



          We need to solve
          $$2sqrt{2x-3}+2x=6$$ or
          $$2x-3+2sqrt{2x-3}+1=4$$ or
          $$left(sqrt{2x-3}+1right)^2=4$$ or
          $$sqrt{2x-3}+1=2$$ or
          $$sqrt{2x-3}=1$$ or $$x=2.$$






          share|cite|improve this answer





















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            5 Answers
            5






            active

            oldest

            votes








            5 Answers
            5






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            6














            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




























              6














              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


























                6












                6








                6






                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 1 hour ago

























                answered 1 hour ago









                KM101

                4,603418




                4,603418























                    8














                    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


























                      8














                      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
























                        8












                        8








                        8






                        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 1 hour ago









                        Siong Thye Goh

                        99k1464116




                        99k1464116























                            1














                            Squaring both sides of an equation can introduce extraneous solutions. Thus it is necessary when doing so to check your answer.



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






                            share|cite|improve this answer


























                              1














                              Squaring both sides of an equation can introduce extraneous solutions. Thus it is necessary when doing so to check your answer.



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






                              share|cite|improve this answer
























                                1












                                1








                                1






                                Squaring both sides of an equation can introduce extraneous solutions. 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. 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










                                answered 1 hour ago









                                Chris Custer

                                10.8k3724




                                10.8k3724























                                    0














                                    The initial question is actually:



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



                                    With each algebraic step, if-then logic is used to rephrase the initial question, eventually leading to:



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



                                    If all the logical steps are 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 inverse of the square root function is not the same as 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, this is much harder than just checking every answer.






                                    share|cite|improve this answer


























                                      0














                                      The initial question is actually:



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



                                      With each algebraic step, if-then logic is used to rephrase the initial question, eventually leading to:



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



                                      If all the logical steps are 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 inverse of the square root function is not the same as 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, this is much harder than just checking every answer.






                                      share|cite|improve this answer
























                                        0












                                        0








                                        0






                                        The initial question is actually:



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



                                        With each algebraic step, if-then logic is used to rephrase the initial question, eventually leading to:



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



                                        If all the logical steps are 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 inverse of the square root function is not the same as 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, this is much harder than just checking every answer.






                                        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 algebraic step, if-then logic is used to rephrase the initial question, eventually leading to:



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



                                        If all the logical steps are 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 inverse of the square root function is not the same as 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, this is much harder than just checking every answer.







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered 1 hour ago









                                        David Diaz

                                        933420




                                        933420























                                            0














                                            We can use also the following way.



                                            We need to solve
                                            $$2sqrt{2x-3}+2x=6$$ or
                                            $$2x-3+2sqrt{2x-3}+1=4$$ or
                                            $$left(sqrt{2x-3}+1right)^2=4$$ or
                                            $$sqrt{2x-3}+1=2$$ or
                                            $$sqrt{2x-3}=1$$ or $$x=2.$$






                                            share|cite|improve this answer


























                                              0














                                              We can use also the following way.



                                              We need to solve
                                              $$2sqrt{2x-3}+2x=6$$ or
                                              $$2x-3+2sqrt{2x-3}+1=4$$ or
                                              $$left(sqrt{2x-3}+1right)^2=4$$ or
                                              $$sqrt{2x-3}+1=2$$ or
                                              $$sqrt{2x-3}=1$$ or $$x=2.$$






                                              share|cite|improve this answer
























                                                0












                                                0








                                                0






                                                We can use also the following way.



                                                We need to solve
                                                $$2sqrt{2x-3}+2x=6$$ or
                                                $$2x-3+2sqrt{2x-3}+1=4$$ or
                                                $$left(sqrt{2x-3}+1right)^2=4$$ or
                                                $$sqrt{2x-3}+1=2$$ or
                                                $$sqrt{2x-3}=1$$ or $$x=2.$$






                                                share|cite|improve this answer












                                                We can use also the following way.



                                                We need to solve
                                                $$2sqrt{2x-3}+2x=6$$ or
                                                $$2x-3+2sqrt{2x-3}+1=4$$ or
                                                $$left(sqrt{2x-3}+1right)^2=4$$ or
                                                $$sqrt{2x-3}+1=2$$ or
                                                $$sqrt{2x-3}=1$$ or $$x=2.$$







                                                share|cite|improve this answer












                                                share|cite|improve this answer



                                                share|cite|improve this answer










                                                answered 25 mins ago









                                                Michael Rozenberg

                                                95.7k1588184




                                                95.7k1588184






























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