Find the sum of $x_1+x_2+x_3$ of intercept points












2












$begingroup$


Suppose that the straight line $L$ meets the curve $y=3x^3-15x^2+7x-8$ in three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$. Then $x_1+x_2+x_3=?$



A) 3 $quad$ B) 4 $quad$ C) 5 $quad$ D) 6 $quad$ E) 7



At the beginning, my main idea is to use Vieta's theorem $x_1+x_2+x_3=-frac{b}{a}$, and find the answer is $5$, it is correct if $y=0$. But when I use software to draw the grapic of $y=3x^3-15x^2+7x-8$ and a line intercept at three points,I also changed the coefficient of the straight line and use calculator did the summation,still can get $x_1+x_2+x_3=5$, I want to know the general solution of this question






contest-math







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    $begingroup$


    Suppose that the straight line $L$ meets the curve $y=3x^3-15x^2+7x-8$ in three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$. Then $x_1+x_2+x_3=?$



    A) 3 $quad$ B) 4 $quad$ C) 5 $quad$ D) 6 $quad$ E) 7



    At the beginning, my main idea is to use Vieta's theorem $x_1+x_2+x_3=-frac{b}{a}$, and find the answer is $5$, it is correct if $y=0$. But when I use software to draw the grapic of $y=3x^3-15x^2+7x-8$ and a line intercept at three points,I also changed the coefficient of the straight line and use calculator did the summation,still can get $x_1+x_2+x_3=5$, I want to know the general solution of this question






    contest-math







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Suppose that the straight line $L$ meets the curve $y=3x^3-15x^2+7x-8$ in three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$. Then $x_1+x_2+x_3=?$



      A) 3 $quad$ B) 4 $quad$ C) 5 $quad$ D) 6 $quad$ E) 7



      At the beginning, my main idea is to use Vieta's theorem $x_1+x_2+x_3=-frac{b}{a}$, and find the answer is $5$, it is correct if $y=0$. But when I use software to draw the grapic of $y=3x^3-15x^2+7x-8$ and a line intercept at three points,I also changed the coefficient of the straight line and use calculator did the summation,still can get $x_1+x_2+x_3=5$, I want to know the general solution of this question






      contest-math







      share|cite|improve this question











      $endgroup$




      Suppose that the straight line $L$ meets the curve $y=3x^3-15x^2+7x-8$ in three points $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$. Then $x_1+x_2+x_3=?$



      A) 3 $quad$ B) 4 $quad$ C) 5 $quad$ D) 6 $quad$ E) 7



      At the beginning, my main idea is to use Vieta's theorem $x_1+x_2+x_3=-frac{b}{a}$, and find the answer is $5$, it is correct if $y=0$. But when I use software to draw the grapic of $y=3x^3-15x^2+7x-8$ and a line intercept at three points,I also changed the coefficient of the straight line and use calculator did the summation,still can get $x_1+x_2+x_3=5$, I want to know the general solution of this question






      contest-math




      contest-math






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      share|cite|improve this question













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      edited Jan 24 at 6:33







      nar

















      asked Jan 24 at 5:51









      narnar

      112




      112






















          2 Answers
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          active

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          3












          $begingroup$

          Let $y=ax+b$ the equation of the straight line $L$. Then $x_1,x_2,x_3$ are solutions of the equation



          $3x^3-15x^2+(7-a)x-8-b=0$.



          Then Vieta says: $x_1+x_2+x_3= - frac{-15}{3}=5.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            All right I didn't notice that, Let $3x^3-15x^2+7x-8=ax+b$, then move $ax+b$ to the left and simplify after that we can use Viet's theorem. Thank you for the idea
            $endgroup$
            – nar
            Jan 24 at 6:16



















          2












          $begingroup$

          If polynomial $p(x) = ax^n + bx^{n-1} + ...$ has roots $x_i$, then $x_1+dots+x_n=-frac{b}{a}$, as you noted.



          Now, though, you don't want $p(x)=0$, you want $p(x)=mx+b$.



          So, to use Vieta's theorem, you need to define a new polynomial:



          Let $q(x)=p(x)-mx-b$. Then, your $x_i$ are the roots of $q(x)$, Vieta's theorem applies, and (because $p$ is cubic or higher), $p$ and $q$ have the same coefficient of $x^n$ and $x^{n-1}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Micheal Hartley Yes, thank you for the solution. I got the idea
            $endgroup$
            – nar
            Jan 24 at 6:18










          • $begingroup$
            @nar and @ Michael Hartley: Viet is Vietnam, Viet is not Vieta....
            $endgroup$
            – Fred
            Jan 24 at 6:26











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Let $y=ax+b$ the equation of the straight line $L$. Then $x_1,x_2,x_3$ are solutions of the equation



          $3x^3-15x^2+(7-a)x-8-b=0$.



          Then Vieta says: $x_1+x_2+x_3= - frac{-15}{3}=5.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            All right I didn't notice that, Let $3x^3-15x^2+7x-8=ax+b$, then move $ax+b$ to the left and simplify after that we can use Viet's theorem. Thank you for the idea
            $endgroup$
            – nar
            Jan 24 at 6:16
















          3












          $begingroup$

          Let $y=ax+b$ the equation of the straight line $L$. Then $x_1,x_2,x_3$ are solutions of the equation



          $3x^3-15x^2+(7-a)x-8-b=0$.



          Then Vieta says: $x_1+x_2+x_3= - frac{-15}{3}=5.$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            All right I didn't notice that, Let $3x^3-15x^2+7x-8=ax+b$, then move $ax+b$ to the left and simplify after that we can use Viet's theorem. Thank you for the idea
            $endgroup$
            – nar
            Jan 24 at 6:16














          3












          3








          3





          $begingroup$

          Let $y=ax+b$ the equation of the straight line $L$. Then $x_1,x_2,x_3$ are solutions of the equation



          $3x^3-15x^2+(7-a)x-8-b=0$.



          Then Vieta says: $x_1+x_2+x_3= - frac{-15}{3}=5.$






          share|cite|improve this answer









          $endgroup$



          Let $y=ax+b$ the equation of the straight line $L$. Then $x_1,x_2,x_3$ are solutions of the equation



          $3x^3-15x^2+(7-a)x-8-b=0$.



          Then Vieta says: $x_1+x_2+x_3= - frac{-15}{3}=5.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 24 at 6:04









          FredFred

          48.3k1849




          48.3k1849












          • $begingroup$
            All right I didn't notice that, Let $3x^3-15x^2+7x-8=ax+b$, then move $ax+b$ to the left and simplify after that we can use Viet's theorem. Thank you for the idea
            $endgroup$
            – nar
            Jan 24 at 6:16


















          • $begingroup$
            All right I didn't notice that, Let $3x^3-15x^2+7x-8=ax+b$, then move $ax+b$ to the left and simplify after that we can use Viet's theorem. Thank you for the idea
            $endgroup$
            – nar
            Jan 24 at 6:16
















          $begingroup$
          All right I didn't notice that, Let $3x^3-15x^2+7x-8=ax+b$, then move $ax+b$ to the left and simplify after that we can use Viet's theorem. Thank you for the idea
          $endgroup$
          – nar
          Jan 24 at 6:16




          $begingroup$
          All right I didn't notice that, Let $3x^3-15x^2+7x-8=ax+b$, then move $ax+b$ to the left and simplify after that we can use Viet's theorem. Thank you for the idea
          $endgroup$
          – nar
          Jan 24 at 6:16











          2












          $begingroup$

          If polynomial $p(x) = ax^n + bx^{n-1} + ...$ has roots $x_i$, then $x_1+dots+x_n=-frac{b}{a}$, as you noted.



          Now, though, you don't want $p(x)=0$, you want $p(x)=mx+b$.



          So, to use Vieta's theorem, you need to define a new polynomial:



          Let $q(x)=p(x)-mx-b$. Then, your $x_i$ are the roots of $q(x)$, Vieta's theorem applies, and (because $p$ is cubic or higher), $p$ and $q$ have the same coefficient of $x^n$ and $x^{n-1}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Micheal Hartley Yes, thank you for the solution. I got the idea
            $endgroup$
            – nar
            Jan 24 at 6:18










          • $begingroup$
            @nar and @ Michael Hartley: Viet is Vietnam, Viet is not Vieta....
            $endgroup$
            – Fred
            Jan 24 at 6:26
















          2












          $begingroup$

          If polynomial $p(x) = ax^n + bx^{n-1} + ...$ has roots $x_i$, then $x_1+dots+x_n=-frac{b}{a}$, as you noted.



          Now, though, you don't want $p(x)=0$, you want $p(x)=mx+b$.



          So, to use Vieta's theorem, you need to define a new polynomial:



          Let $q(x)=p(x)-mx-b$. Then, your $x_i$ are the roots of $q(x)$, Vieta's theorem applies, and (because $p$ is cubic or higher), $p$ and $q$ have the same coefficient of $x^n$ and $x^{n-1}$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            @Micheal Hartley Yes, thank you for the solution. I got the idea
            $endgroup$
            – nar
            Jan 24 at 6:18










          • $begingroup$
            @nar and @ Michael Hartley: Viet is Vietnam, Viet is not Vieta....
            $endgroup$
            – Fred
            Jan 24 at 6:26














          2












          2








          2





          $begingroup$

          If polynomial $p(x) = ax^n + bx^{n-1} + ...$ has roots $x_i$, then $x_1+dots+x_n=-frac{b}{a}$, as you noted.



          Now, though, you don't want $p(x)=0$, you want $p(x)=mx+b$.



          So, to use Vieta's theorem, you need to define a new polynomial:



          Let $q(x)=p(x)-mx-b$. Then, your $x_i$ are the roots of $q(x)$, Vieta's theorem applies, and (because $p$ is cubic or higher), $p$ and $q$ have the same coefficient of $x^n$ and $x^{n-1}$.






          share|cite|improve this answer











          $endgroup$



          If polynomial $p(x) = ax^n + bx^{n-1} + ...$ has roots $x_i$, then $x_1+dots+x_n=-frac{b}{a}$, as you noted.



          Now, though, you don't want $p(x)=0$, you want $p(x)=mx+b$.



          So, to use Vieta's theorem, you need to define a new polynomial:



          Let $q(x)=p(x)-mx-b$. Then, your $x_i$ are the roots of $q(x)$, Vieta's theorem applies, and (because $p$ is cubic or higher), $p$ and $q$ have the same coefficient of $x^n$ and $x^{n-1}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 25 at 8:20

























          answered Jan 24 at 6:05









          Michael HartleyMichael Hartley

          1,10555




          1,10555












          • $begingroup$
            @Micheal Hartley Yes, thank you for the solution. I got the idea
            $endgroup$
            – nar
            Jan 24 at 6:18










          • $begingroup$
            @nar and @ Michael Hartley: Viet is Vietnam, Viet is not Vieta....
            $endgroup$
            – Fred
            Jan 24 at 6:26


















          • $begingroup$
            @Micheal Hartley Yes, thank you for the solution. I got the idea
            $endgroup$
            – nar
            Jan 24 at 6:18










          • $begingroup$
            @nar and @ Michael Hartley: Viet is Vietnam, Viet is not Vieta....
            $endgroup$
            – Fred
            Jan 24 at 6:26
















          $begingroup$
          @Micheal Hartley Yes, thank you for the solution. I got the idea
          $endgroup$
          – nar
          Jan 24 at 6:18




          $begingroup$
          @Micheal Hartley Yes, thank you for the solution. I got the idea
          $endgroup$
          – nar
          Jan 24 at 6:18












          $begingroup$
          @nar and @ Michael Hartley: Viet is Vietnam, Viet is not Vieta....
          $endgroup$
          – Fred
          Jan 24 at 6:26




          $begingroup$
          @nar and @ Michael Hartley: Viet is Vietnam, Viet is not Vieta....
          $endgroup$
          – Fred
          Jan 24 at 6:26


















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