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
    yesterday






  • 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
    yesterday








  • 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
    yesterday








  • 1




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






  • 1




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
















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
    yesterday






  • 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
    yesterday








  • 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
    yesterday








  • 1




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






  • 1




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














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 7 hours ago







Beginner

















asked yesterday









BeginnerBeginner

1719




1719








  • 7




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






  • 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
    yesterday








  • 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
    yesterday








  • 1




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






  • 1




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














  • 7




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






  • 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
    yesterday








  • 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
    yesterday








  • 1




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






  • 1




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








7




7




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




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




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
yesterday






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
yesterday






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
yesterday






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
yesterday






1




1




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




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




1




1




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




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










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
    15 hours ago










  • @DavidZ, excellent point!
    – Barry Cipra
    6 hours 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
      yesterday










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






    • 1




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






    • 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
      yesterday






    • 1




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










    • @DavidZ, excellent point!
      – Barry Cipra
      6 hours 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
      15 hours ago










    • @DavidZ, excellent point!
      – Barry Cipra
      6 hours 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 yesterday









    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
      15 hours ago










    • @DavidZ, excellent point!
      – Barry Cipra
      6 hours 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
      15 hours ago










    • @DavidZ, excellent point!
      – Barry Cipra
      6 hours 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
    15 hours ago




    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
    15 hours ago












    @DavidZ, excellent point!
    – Barry Cipra
    6 hours ago




    @DavidZ, excellent point!
    – Barry Cipra
    6 hours 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 yesterday









        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
              yesterday










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






            • 1




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






            • 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
              yesterday






            • 1




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










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






            • 1




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






            • 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
              yesterday






            • 1




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

























            answered yesterday









            Michael BurrMichael Burr

            26.6k23262




            26.6k23262












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










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






            • 1




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






            • 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
              yesterday






            • 1




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




















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










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






            • 1




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






            • 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
              yesterday






            • 1




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


















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




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












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




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




            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
            yesterday




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




            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
            yesterday




            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
            yesterday




            1




            1




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






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




















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