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
    2 days ago






  • 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
    2 days ago








  • 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
    2 days ago








  • 1




    @String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
    – Beginner
    2 days ago






  • 1




    @Beginner: Yes, and then you must find integer solutions and count values of $b$.
    – String
    2 days ago
















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
    2 days ago






  • 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
    2 days ago








  • 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
    2 days ago








  • 1




    @String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
    – Beginner
    2 days ago






  • 1




    @Beginner: Yes, and then you must find integer solutions and count values of $b$.
    – String
    2 days ago














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 yesterday







Beginner

















asked 2 days ago









BeginnerBeginner

30510




30510








  • 7




    I think there is a problem with the question. Do you have the source of it?
    – Dr. Sonnhard Graubner
    2 days ago






  • 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
    2 days ago








  • 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
    2 days ago








  • 1




    @String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
    – Beginner
    2 days ago






  • 1




    @Beginner: Yes, and then you must find integer solutions and count values of $b$.
    – String
    2 days ago














  • 7




    I think there is a problem with the question. Do you have the source of it?
    – Dr. Sonnhard Graubner
    2 days ago






  • 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
    2 days ago








  • 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
    2 days ago








  • 1




    @String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
    – Beginner
    2 days ago






  • 1




    @Beginner: Yes, and then you must find integer solutions and count values of $b$.
    – String
    2 days ago








7




7




I think there is a problem with the question. Do you have the source of it?
– Dr. Sonnhard Graubner
2 days ago




I think there is a problem with the question. Do you have the source of it?
– Dr. Sonnhard Graubner
2 days ago




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
2 days ago






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
2 days ago






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
2 days ago






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
2 days ago






1




1




@String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
– Beginner
2 days ago




@String $6+a+b+c=0$, and $6a+3-bc=0$ ..??
– Beginner
2 days ago




1




1




@Beginner: Yes, and then you must find integer solutions and count values of $b$.
– String
2 days ago




@Beginner: Yes, and then you must find integer solutions and count values of $b$.
– String
2 days ago










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
    yesterday










  • @DavidZ, excellent point!
    – Barry Cipra
    yesterday



















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
      2 days ago










    • @JohnOmielan Thanks, that's what I get for answering on my phone.
      – Michael Burr
      2 days ago






    • 1




      @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
      – String
      2 days ago






    • 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
      2 days ago






    • 1




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













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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
      yesterday










    • @DavidZ, excellent point!
      – Barry Cipra
      yesterday
















    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
      yesterday










    • @DavidZ, excellent point!
      – Barry Cipra
      yesterday














    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 2 days ago









    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
      yesterday










    • @DavidZ, excellent point!
      – Barry Cipra
      yesterday














    • 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
      yesterday










    • @DavidZ, excellent point!
      – Barry Cipra
      yesterday








    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
    yesterday




    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
    yesterday












    @DavidZ, excellent point!
    – Barry Cipra
    yesterday




    @DavidZ, excellent point!
    – Barry Cipra
    yesterday











    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 2 days ago









        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
              2 days ago










            • @JohnOmielan Thanks, that's what I get for answering on my phone.
              – Michael Burr
              2 days ago






            • 1




              @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
              – String
              2 days ago






            • 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
              2 days ago






            • 1




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


















            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
              2 days ago










            • @JohnOmielan Thanks, that's what I get for answering on my phone.
              – Michael Burr
              2 days ago






            • 1




              @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
              – String
              2 days ago






            • 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
              2 days ago






            • 1




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
















            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 2 days ago

























            answered 2 days ago









            Michael BurrMichael Burr

            26.6k23262




            26.6k23262












            • You forgot a "$+ 3$" on the left side of your final equation.
              – John Omielan
              2 days ago










            • @JohnOmielan Thanks, that's what I get for answering on my phone.
              – Michael Burr
              2 days ago






            • 1




              @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
              – String
              2 days ago






            • 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
              2 days ago






            • 1




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




















            • You forgot a "$+ 3$" on the left side of your final equation.
              – John Omielan
              2 days ago










            • @JohnOmielan Thanks, that's what I get for answering on my phone.
              – Michael Burr
              2 days ago






            • 1




              @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
              – String
              2 days ago






            • 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
              2 days ago






            • 1




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


















            You forgot a "$+ 3$" on the left side of your final equation.
            – John Omielan
            2 days ago




            You forgot a "$+ 3$" on the left side of your final equation.
            – John Omielan
            2 days ago












            @JohnOmielan Thanks, that's what I get for answering on my phone.
            – Michael Burr
            2 days ago




            @JohnOmielan Thanks, that's what I get for answering on my phone.
            – Michael Burr
            2 days ago




            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
            2 days ago




            @Beginner: I would have put it as "Given some $a,b,cin mathbb Z$ the following equality holds for all $xinmathbb R$ ..."
            – String
            2 days ago




            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
            2 days ago




            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
            2 days ago




            1




            1




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






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




















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