What is the mathematical meaning of this question?












5















$a,b,c inmathbb{Z}$ and $xinmathbb{R}$, then the following expression is always true:



$$(x-a)(x-6)+3=(x+b)(x+c)$$



Find the sum of all possible values of $b$.



A) $-8$



B) $-12$



C) $-14$



D) $-24$



E) $-16$




I didn't understand what is the meaning of "...is always true".



Even though I can't understand the question, I wrote these:



$$(x-a)(x-6)+3=(x+b)(x+c) Rightarrow x=frac{6a-bc+3}{6+a+b+c}$$



Here, $b$ can take an infinite number of values. Or do I miss something? For example, let random values $a=100,b=50,c=3$ then $x=frac {151}{53}$.



Is there a problem with the question?






algebra-precalculus contest-math problem-solving means







share|cite|improve this question




















  • 7




    I think there is a problem with the question. Do you have the source of it?
    – Dr. Sonnhard Graubner
    Jan 6 at 12:00






  • 3




    Hint: Equate the coefficients of $x$. The "are always true" means that the equality holds for every $x$ you choose. The problem with your calculation for $x$ is that the denominator of that fraction vanishes.
    – Michael Burr
    Jan 6 at 12:03








  • 2




    Here are a few details: $$x^2-(a+6)x+6a+3=x^2+(b+c)x+bc$$ and for this to be true for ALL values of $x$ we must have equal coefficients on both sides.
    – String
    Jan 6 at 12:09








  • 1




    @String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
    – Beginner
    Jan 6 at 12:13






  • 1




    @Beginner: Yes, and then you must find integer solutions and count values of $b$.
    – String
    Jan 6 at 12:14
















5















$a,b,c inmathbb{Z}$ and $xinmathbb{R}$, then the following expression is always true:



$$(x-a)(x-6)+3=(x+b)(x+c)$$



Find the sum of all possible values of $b$.



A) $-8$



B) $-12$



C) $-14$



D) $-24$



E) $-16$




I didn't understand what is the meaning of "...is always true".



Even though I can't understand the question, I wrote these:



$$(x-a)(x-6)+3=(x+b)(x+c) Rightarrow x=frac{6a-bc+3}{6+a+b+c}$$



Here, $b$ can take an infinite number of values. Or do I miss something? For example, let random values $a=100,b=50,c=3$ then $x=frac {151}{53}$.



Is there a problem with the question?






algebra-precalculus contest-math problem-solving means







share|cite|improve this question




















  • 7




    I think there is a problem with the question. Do you have the source of it?
    – Dr. Sonnhard Graubner
    Jan 6 at 12:00






  • 3




    Hint: Equate the coefficients of $x$. The "are always true" means that the equality holds for every $x$ you choose. The problem with your calculation for $x$ is that the denominator of that fraction vanishes.
    – Michael Burr
    Jan 6 at 12:03








  • 2




    Here are a few details: $$x^2-(a+6)x+6a+3=x^2+(b+c)x+bc$$ and for this to be true for ALL values of $x$ we must have equal coefficients on both sides.
    – String
    Jan 6 at 12:09








  • 1




    @String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
    – Beginner
    Jan 6 at 12:13






  • 1




    @Beginner: Yes, and then you must find integer solutions and count values of $b$.
    – String
    Jan 6 at 12:14














5












5








5


2






$a,b,c inmathbb{Z}$ and $xinmathbb{R}$, then the following expression is always true:



$$(x-a)(x-6)+3=(x+b)(x+c)$$



Find the sum of all possible values of $b$.



A) $-8$



B) $-12$



C) $-14$



D) $-24$



E) $-16$




I didn't understand what is the meaning of "...is always true".



Even though I can't understand the question, I wrote these:



$$(x-a)(x-6)+3=(x+b)(x+c) Rightarrow x=frac{6a-bc+3}{6+a+b+c}$$



Here, $b$ can take an infinite number of values. Or do I miss something? For example, let random values $a=100,b=50,c=3$ then $x=frac {151}{53}$.



Is there a problem with the question?






algebra-precalculus contest-math problem-solving means







share|cite|improve this question
















$a,b,c inmathbb{Z}$ and $xinmathbb{R}$, then the following expression is always true:



$$(x-a)(x-6)+3=(x+b)(x+c)$$



Find the sum of all possible values of $b$.



A) $-8$



B) $-12$



C) $-14$



D) $-24$



E) $-16$




I didn't understand what is the meaning of "...is always true".



Even though I can't understand the question, I wrote these:



$$(x-a)(x-6)+3=(x+b)(x+c) Rightarrow x=frac{6a-bc+3}{6+a+b+c}$$



Here, $b$ can take an infinite number of values. Or do I miss something? For example, let random values $a=100,b=50,c=3$ then $x=frac {151}{53}$.



Is there a problem with the question?






algebra-precalculus contest-math problem-solving means




algebra-precalculus contest-math problem-solving means






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago







Beginner

















asked Jan 6 at 11:51









BeginnerBeginner

30510




30510








  • 7




    I think there is a problem with the question. Do you have the source of it?
    – Dr. Sonnhard Graubner
    Jan 6 at 12:00






  • 3




    Hint: Equate the coefficients of $x$. The "are always true" means that the equality holds for every $x$ you choose. The problem with your calculation for $x$ is that the denominator of that fraction vanishes.
    – Michael Burr
    Jan 6 at 12:03








  • 2




    Here are a few details: $$x^2-(a+6)x+6a+3=x^2+(b+c)x+bc$$ and for this to be true for ALL values of $x$ we must have equal coefficients on both sides.
    – String
    Jan 6 at 12:09








  • 1




    @String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
    – Beginner
    Jan 6 at 12:13






  • 1




    @Beginner: Yes, and then you must find integer solutions and count values of $b$.
    – String
    Jan 6 at 12:14














  • 7




    I think there is a problem with the question. Do you have the source of it?
    – Dr. Sonnhard Graubner
    Jan 6 at 12:00






  • 3




    Hint: Equate the coefficients of $x$. The "are always true" means that the equality holds for every $x$ you choose. The problem with your calculation for $x$ is that the denominator of that fraction vanishes.
    – Michael Burr
    Jan 6 at 12:03








  • 2




    Here are a few details: $$x^2-(a+6)x+6a+3=x^2+(b+c)x+bc$$ and for this to be true for ALL values of $x$ we must have equal coefficients on both sides.
    – String
    Jan 6 at 12:09








  • 1




    @String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
    – Beginner
    Jan 6 at 12:13






  • 1




    @Beginner: Yes, and then you must find integer solutions and count values of $b$.
    – String
    Jan 6 at 12:14








7




7




I think there is a problem with the question. Do you have the source of it?
– Dr. Sonnhard Graubner
Jan 6 at 12:00




I think there is a problem with the question. Do you have the source of it?
– Dr. Sonnhard Graubner
Jan 6 at 12:00




3




3




Hint: Equate the coefficients of $x$. The "are always true" means that the equality holds for every $x$ you choose. The problem with your calculation for $x$ is that the denominator of that fraction vanishes.
– Michael Burr
Jan 6 at 12:03






Hint: Equate the coefficients of $x$. The "are always true" means that the equality holds for every $x$ you choose. The problem with your calculation for $x$ is that the denominator of that fraction vanishes.
– Michael Burr
Jan 6 at 12:03






2




2




Here are a few details: $$x^2-(a+6)x+6a+3=x^2+(b+c)x+bc$$ and for this to be true for ALL values of $x$ we must have equal coefficients on both sides.
– String
Jan 6 at 12:09






Here are a few details: $$x^2-(a+6)x+6a+3=x^2+(b+c)x+bc$$ and for this to be true for ALL values of $x$ we must have equal coefficients on both sides.
– String
Jan 6 at 12:09






1




1




@String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
– Beginner
Jan 6 at 12:13




@String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
– Beginner
Jan 6 at 12:13




1




1




@Beginner: Yes, and then you must find integer solutions and count values of $b$.
– String
Jan 6 at 12:14




@Beginner: Yes, and then you must find integer solutions and count values of $b$.
– String
Jan 6 at 12:14










3 Answers
3






active

oldest

votes


















20














To answer the explicit question, "Is there a problem with the question?," the answer is Yes, it's worded in a weird, nonsensical way. (I think this is why Dr. Sonnhard Graubner left a comment asking for the question's source: was it reproduced verbatim, or did the OP paraphrase the problem?) A better version would be something like this:




Consider the set of triples
$(a,b,c)inmathbb{Z^3}$ for which the equation



$$(x-a)(x-6)+3=(x+b)(x+c)$$



holds for all $xinmathbb{R}$. Find the sum of all the $b$'s among
these triples.







share|cite|improve this answer

















  • 8




    Just a little nitpick, I think this is not strictly equivalent to the wording in the question, since if two triples have the same value of $b$ but differ in $a$ and $c$, the wording in the question would count that value of $b$ only once but your version counts it twice. Of course I don't think that impedes your point at all.
    – David Z
    Jan 7 at 3:45










  • @DavidZ, excellent point!
    – Barry Cipra
    2 days ago



















11














The question is poorly worded. It should read something like this:




$a,b,c$ are integers such that the following equation holds for all
$xinBbb R$:




etc.






share|cite|improve this answer





























    7














    Two polynomials which are always equal over the reals are exactly the same. In this case, since $x$ is allowed to vary, while $a,b,c$ are fixed, these are two polynomials in $x$.



    For them to be equal, the coefficients of $x$ must also be equal. Therefore,
    begin{align}
    -a-6&=b+c\
    6a+3&=bc.
    end{align}
    Now, you can solve for $a$ in the first equation and substitute into the second equation, giving
    $$
    6(-b-c-6)+3=bc.
    $$
    The problem then becomes, for which integers does this equation have a solution?



    If you solve for $b$ here, you'll get a fraction in $c$, which you can study to figure out which integers for $c$ result in integers for $b$.



    The problem with your solution for $x$ is that the denominator of your fraction is zero.






    share|cite|improve this answer























    • You forgot a "$+ 3$" on the left side of your final equation.
      – John Omielan
      Jan 6 at 12:16










    • @JohnOmielan Thanks, that's what I get for answering on my phone.
      – Michael Burr
      Jan 6 at 12:18






    • 1




      @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
      – String
      Jan 6 at 12:22






    • 1




      I agree that expression is not the best word. Equality or statement would be better. Technically, it would be ok because the equality evaluates to true or false, which is an expression.
      – Michael Burr
      Jan 6 at 12:47






    • 1




      The last equation can be rewritten as $(b+6)(c+6) = 3$ which is easier to solve for integers.
      – yper-crazyhat-cubeᵀᴹ
      2 days ago













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    20














    To answer the explicit question, "Is there a problem with the question?," the answer is Yes, it's worded in a weird, nonsensical way. (I think this is why Dr. Sonnhard Graubner left a comment asking for the question's source: was it reproduced verbatim, or did the OP paraphrase the problem?) A better version would be something like this:




    Consider the set of triples
    $(a,b,c)inmathbb{Z^3}$ for which the equation



    $$(x-a)(x-6)+3=(x+b)(x+c)$$



    holds for all $xinmathbb{R}$. Find the sum of all the $b$'s among
    these triples.







    share|cite|improve this answer

















    • 8




      Just a little nitpick, I think this is not strictly equivalent to the wording in the question, since if two triples have the same value of $b$ but differ in $a$ and $c$, the wording in the question would count that value of $b$ only once but your version counts it twice. Of course I don't think that impedes your point at all.
      – David Z
      Jan 7 at 3:45










    • @DavidZ, excellent point!
      – Barry Cipra
      2 days ago
















    20














    To answer the explicit question, "Is there a problem with the question?," the answer is Yes, it's worded in a weird, nonsensical way. (I think this is why Dr. Sonnhard Graubner left a comment asking for the question's source: was it reproduced verbatim, or did the OP paraphrase the problem?) A better version would be something like this:




    Consider the set of triples
    $(a,b,c)inmathbb{Z^3}$ for which the equation



    $$(x-a)(x-6)+3=(x+b)(x+c)$$



    holds for all $xinmathbb{R}$. Find the sum of all the $b$'s among
    these triples.







    share|cite|improve this answer

















    • 8




      Just a little nitpick, I think this is not strictly equivalent to the wording in the question, since if two triples have the same value of $b$ but differ in $a$ and $c$, the wording in the question would count that value of $b$ only once but your version counts it twice. Of course I don't think that impedes your point at all.
      – David Z
      Jan 7 at 3:45










    • @DavidZ, excellent point!
      – Barry Cipra
      2 days ago














    20












    20








    20






    To answer the explicit question, "Is there a problem with the question?," the answer is Yes, it's worded in a weird, nonsensical way. (I think this is why Dr. Sonnhard Graubner left a comment asking for the question's source: was it reproduced verbatim, or did the OP paraphrase the problem?) A better version would be something like this:




    Consider the set of triples
    $(a,b,c)inmathbb{Z^3}$ for which the equation



    $$(x-a)(x-6)+3=(x+b)(x+c)$$



    holds for all $xinmathbb{R}$. Find the sum of all the $b$'s among
    these triples.







    share|cite|improve this answer












    To answer the explicit question, "Is there a problem with the question?," the answer is Yes, it's worded in a weird, nonsensical way. (I think this is why Dr. Sonnhard Graubner left a comment asking for the question's source: was it reproduced verbatim, or did the OP paraphrase the problem?) A better version would be something like this:




    Consider the set of triples
    $(a,b,c)inmathbb{Z^3}$ for which the equation



    $$(x-a)(x-6)+3=(x+b)(x+c)$$



    holds for all $xinmathbb{R}$. Find the sum of all the $b$'s among
    these triples.








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 6 at 12:40









    Barry CipraBarry Cipra

    59.3k653125




    59.3k653125








    • 8




      Just a little nitpick, I think this is not strictly equivalent to the wording in the question, since if two triples have the same value of $b$ but differ in $a$ and $c$, the wording in the question would count that value of $b$ only once but your version counts it twice. Of course I don't think that impedes your point at all.
      – David Z
      Jan 7 at 3:45










    • @DavidZ, excellent point!
      – Barry Cipra
      2 days ago














    • 8




      Just a little nitpick, I think this is not strictly equivalent to the wording in the question, since if two triples have the same value of $b$ but differ in $a$ and $c$, the wording in the question would count that value of $b$ only once but your version counts it twice. Of course I don't think that impedes your point at all.
      – David Z
      Jan 7 at 3:45










    • @DavidZ, excellent point!
      – Barry Cipra
      2 days ago








    8




    8




    Just a little nitpick, I think this is not strictly equivalent to the wording in the question, since if two triples have the same value of $b$ but differ in $a$ and $c$, the wording in the question would count that value of $b$ only once but your version counts it twice. Of course I don't think that impedes your point at all.
    – David Z
    Jan 7 at 3:45




    Just a little nitpick, I think this is not strictly equivalent to the wording in the question, since if two triples have the same value of $b$ but differ in $a$ and $c$, the wording in the question would count that value of $b$ only once but your version counts it twice. Of course I don't think that impedes your point at all.
    – David Z
    Jan 7 at 3:45












    @DavidZ, excellent point!
    – Barry Cipra
    2 days ago




    @DavidZ, excellent point!
    – Barry Cipra
    2 days ago











    11














    The question is poorly worded. It should read something like this:




    $a,b,c$ are integers such that the following equation holds for all
    $xinBbb R$:




    etc.






    share|cite|improve this answer


























      11














      The question is poorly worded. It should read something like this:




      $a,b,c$ are integers such that the following equation holds for all
      $xinBbb R$:




      etc.






      share|cite|improve this answer
























        11












        11








        11






        The question is poorly worded. It should read something like this:




        $a,b,c$ are integers such that the following equation holds for all
        $xinBbb R$:




        etc.






        share|cite|improve this answer












        The question is poorly worded. It should read something like this:




        $a,b,c$ are integers such that the following equation holds for all
        $xinBbb R$:




        etc.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 12:27









        TonyKTonyK

        41.7k353133




        41.7k353133























            7














            Two polynomials which are always equal over the reals are exactly the same. In this case, since $x$ is allowed to vary, while $a,b,c$ are fixed, these are two polynomials in $x$.



            For them to be equal, the coefficients of $x$ must also be equal. Therefore,
            begin{align}
            -a-6&=b+c\
            6a+3&=bc.
            end{align}
            Now, you can solve for $a$ in the first equation and substitute into the second equation, giving
            $$
            6(-b-c-6)+3=bc.
            $$
            The problem then becomes, for which integers does this equation have a solution?



            If you solve for $b$ here, you'll get a fraction in $c$, which you can study to figure out which integers for $c$ result in integers for $b$.



            The problem with your solution for $x$ is that the denominator of your fraction is zero.






            share|cite|improve this answer























            • You forgot a "$+ 3$" on the left side of your final equation.
              – John Omielan
              Jan 6 at 12:16










            • @JohnOmielan Thanks, that's what I get for answering on my phone.
              – Michael Burr
              Jan 6 at 12:18






            • 1




              @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
              – String
              Jan 6 at 12:22






            • 1




              I agree that expression is not the best word. Equality or statement would be better. Technically, it would be ok because the equality evaluates to true or false, which is an expression.
              – Michael Burr
              Jan 6 at 12:47






            • 1




              The last equation can be rewritten as $(b+6)(c+6) = 3$ which is easier to solve for integers.
              – yper-crazyhat-cubeᵀᴹ
              2 days ago


















            7














            Two polynomials which are always equal over the reals are exactly the same. In this case, since $x$ is allowed to vary, while $a,b,c$ are fixed, these are two polynomials in $x$.



            For them to be equal, the coefficients of $x$ must also be equal. Therefore,
            begin{align}
            -a-6&=b+c\
            6a+3&=bc.
            end{align}
            Now, you can solve for $a$ in the first equation and substitute into the second equation, giving
            $$
            6(-b-c-6)+3=bc.
            $$
            The problem then becomes, for which integers does this equation have a solution?



            If you solve for $b$ here, you'll get a fraction in $c$, which you can study to figure out which integers for $c$ result in integers for $b$.



            The problem with your solution for $x$ is that the denominator of your fraction is zero.






            share|cite|improve this answer























            • You forgot a "$+ 3$" on the left side of your final equation.
              – John Omielan
              Jan 6 at 12:16










            • @JohnOmielan Thanks, that's what I get for answering on my phone.
              – Michael Burr
              Jan 6 at 12:18






            • 1




              @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
              – String
              Jan 6 at 12:22






            • 1




              I agree that expression is not the best word. Equality or statement would be better. Technically, it would be ok because the equality evaluates to true or false, which is an expression.
              – Michael Burr
              Jan 6 at 12:47






            • 1




              The last equation can be rewritten as $(b+6)(c+6) = 3$ which is easier to solve for integers.
              – yper-crazyhat-cubeᵀᴹ
              2 days ago
















            7












            7








            7






            Two polynomials which are always equal over the reals are exactly the same. In this case, since $x$ is allowed to vary, while $a,b,c$ are fixed, these are two polynomials in $x$.



            For them to be equal, the coefficients of $x$ must also be equal. Therefore,
            begin{align}
            -a-6&=b+c\
            6a+3&=bc.
            end{align}
            Now, you can solve for $a$ in the first equation and substitute into the second equation, giving
            $$
            6(-b-c-6)+3=bc.
            $$
            The problem then becomes, for which integers does this equation have a solution?



            If you solve for $b$ here, you'll get a fraction in $c$, which you can study to figure out which integers for $c$ result in integers for $b$.



            The problem with your solution for $x$ is that the denominator of your fraction is zero.






            share|cite|improve this answer














            Two polynomials which are always equal over the reals are exactly the same. In this case, since $x$ is allowed to vary, while $a,b,c$ are fixed, these are two polynomials in $x$.



            For them to be equal, the coefficients of $x$ must also be equal. Therefore,
            begin{align}
            -a-6&=b+c\
            6a+3&=bc.
            end{align}
            Now, you can solve for $a$ in the first equation and substitute into the second equation, giving
            $$
            6(-b-c-6)+3=bc.
            $$
            The problem then becomes, for which integers does this equation have a solution?



            If you solve for $b$ here, you'll get a fraction in $c$, which you can study to figure out which integers for $c$ result in integers for $b$.



            The problem with your solution for $x$ is that the denominator of your fraction is zero.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 6 at 12:17

























            answered Jan 6 at 12:13









            Michael BurrMichael Burr

            26.6k23262




            26.6k23262












            • You forgot a "$+ 3$" on the left side of your final equation.
              – John Omielan
              Jan 6 at 12:16










            • @JohnOmielan Thanks, that's what I get for answering on my phone.
              – Michael Burr
              Jan 6 at 12:18






            • 1




              @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
              – String
              Jan 6 at 12:22






            • 1




              I agree that expression is not the best word. Equality or statement would be better. Technically, it would be ok because the equality evaluates to true or false, which is an expression.
              – Michael Burr
              Jan 6 at 12:47






            • 1




              The last equation can be rewritten as $(b+6)(c+6) = 3$ which is easier to solve for integers.
              – yper-crazyhat-cubeᵀᴹ
              2 days ago




















            • You forgot a "$+ 3$" on the left side of your final equation.
              – John Omielan
              Jan 6 at 12:16










            • @JohnOmielan Thanks, that's what I get for answering on my phone.
              – Michael Burr
              Jan 6 at 12:18






            • 1




              @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
              – String
              Jan 6 at 12:22






            • 1




              I agree that expression is not the best word. Equality or statement would be better. Technically, it would be ok because the equality evaluates to true or false, which is an expression.
              – Michael Burr
              Jan 6 at 12:47






            • 1




              The last equation can be rewritten as $(b+6)(c+6) = 3$ which is easier to solve for integers.
              – yper-crazyhat-cubeᵀᴹ
              2 days ago


















            You forgot a "$+ 3$" on the left side of your final equation.
            – John Omielan
            Jan 6 at 12:16




            You forgot a "$+ 3$" on the left side of your final equation.
            – John Omielan
            Jan 6 at 12:16












            @JohnOmielan Thanks, that's what I get for answering on my phone.
            – Michael Burr
            Jan 6 at 12:18




            @JohnOmielan Thanks, that's what I get for answering on my phone.
            – Michael Burr
            Jan 6 at 12:18




            1




            1




            @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
            – String
            Jan 6 at 12:22




            @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
            – String
            Jan 6 at 12:22




            1




            1




            I agree that expression is not the best word. Equality or statement would be better. Technically, it would be ok because the equality evaluates to true or false, which is an expression.
            – Michael Burr
            Jan 6 at 12:47




            I agree that expression is not the best word. Equality or statement would be better. Technically, it would be ok because the equality evaluates to true or false, which is an expression.
            – Michael Burr
            Jan 6 at 12:47




            1




            1




            The last equation can be rewritten as $(b+6)(c+6) = 3$ which is easier to solve for integers.
            – yper-crazyhat-cubeᵀᴹ
            2 days ago






            The last equation can be rewritten as $(b+6)(c+6) = 3$ which is easier to solve for integers.
            – yper-crazyhat-cubeᵀᴹ
            2 days ago




















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