Polynomial equation, cannot solve for $x$












2















$$3x^2-4x-4+x^3=x^3+2x+2$$




This boils down to (I think):



$$3x^2 - 6x - 6 = 0$$



I'm trying to solve for $x$ using the polynomial equation:
$$begin{align}
x &= frac{-bpmsqrt{b^2-4ac}}{2a} \
a &= 3 \[0.2ex]
b &= -6 \
c &= -6
end{align}$$



My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3}$. I'm unable to replicate this.



Here's my work:
$$begin{align}
x&=frac{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{36-72}}{2a}
end{align}$$



Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.



Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into



$$3x^2-6x-6=0$$



How can I arrive at solutions for $x$ where $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3},$ per my textbook's given solutions?






polynomials







share|cite|improve this question




















  • 1




    discriminant will be $36+72$, that's your mistake
    – Vasya
    Sep 17 '18 at 15:34
















2















$$3x^2-4x-4+x^3=x^3+2x+2$$




This boils down to (I think):



$$3x^2 - 6x - 6 = 0$$



I'm trying to solve for $x$ using the polynomial equation:
$$begin{align}
x &= frac{-bpmsqrt{b^2-4ac}}{2a} \
a &= 3 \[0.2ex]
b &= -6 \
c &= -6
end{align}$$



My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3}$. I'm unable to replicate this.



Here's my work:
$$begin{align}
x&=frac{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{36-72}}{2a}
end{align}$$



Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.



Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into



$$3x^2-6x-6=0$$



How can I arrive at solutions for $x$ where $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3},$ per my textbook's given solutions?






polynomials







share|cite|improve this question




















  • 1




    discriminant will be $36+72$, that's your mistake
    – Vasya
    Sep 17 '18 at 15:34














2












2








2








$$3x^2-4x-4+x^3=x^3+2x+2$$




This boils down to (I think):



$$3x^2 - 6x - 6 = 0$$



I'm trying to solve for $x$ using the polynomial equation:
$$begin{align}
x &= frac{-bpmsqrt{b^2-4ac}}{2a} \
a &= 3 \[0.2ex]
b &= -6 \
c &= -6
end{align}$$



My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3}$. I'm unable to replicate this.



Here's my work:
$$begin{align}
x&=frac{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{36-72}}{2a}
end{align}$$



Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.



Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into



$$3x^2-6x-6=0$$



How can I arrive at solutions for $x$ where $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3},$ per my textbook's given solutions?






polynomials







share|cite|improve this question
















$$3x^2-4x-4+x^3=x^3+2x+2$$




This boils down to (I think):



$$3x^2 - 6x - 6 = 0$$



I'm trying to solve for $x$ using the polynomial equation:
$$begin{align}
x &= frac{-bpmsqrt{b^2-4ac}}{2a} \
a &= 3 \[0.2ex]
b &= -6 \
c &= -6
end{align}$$



My textbook tells me that the solutions for the positive and negative versions of $x$ are $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3}$. I'm unable to replicate this.



Here's my work:
$$begin{align}
x&=frac{-bpmsqrt{b^2-4ac}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{(-6)^2-4(3)(-6)}}{2a} \[0.8ex]
x&=frac{-6pmsqrt{36-72}}{2a}
end{align}$$



Since $(36 - 72) < 0$, I'm unable to take the square root so I do not see how the equation is solvable.



Unless I boiled down the equation into the wrong form with the original.
I turned:
$$3x^2-4x-4+x^3=x^3+2x+2$$
into



$$3x^2-6x-6=0$$



How can I arrive at solutions for $x$ where $x = 1 + sqrt{3}$ and $x = 1 - sqrt{3},$ per my textbook's given solutions?






polynomials




polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 21:01









Robert Howard

1,9161822




1,9161822










asked Sep 17 '18 at 15:16









Doug Fir

1837




1837








  • 1




    discriminant will be $36+72$, that's your mistake
    – Vasya
    Sep 17 '18 at 15:34














  • 1




    discriminant will be $36+72$, that's your mistake
    – Vasya
    Sep 17 '18 at 15:34








1




1




discriminant will be $36+72$, that's your mistake
– Vasya
Sep 17 '18 at 15:34




discriminant will be $36+72$, that's your mistake
– Vasya
Sep 17 '18 at 15:34










2 Answers
2






active

oldest

votes


















2














so we start off with:
$$x^3+3x^2-4x-4=x^3+2x+2$$
which we can re-write as:
$$3x^2-6x-6=0$$
so we know that:
$$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$



The mistake you made was $36-72$ instead of $36+72$






share|cite|improve this answer





















  • Thanks but would 108 not be 114 since the initial 6 +-?
    – Doug Fir
    Sep 17 '18 at 15:49






  • 1




    that is not inside the squareroot so we do not include it
    – Henry Lee
    Sep 17 '18 at 15:51










  • mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
    – Henry Lee
    Sep 17 '18 at 15:52



















1














$$x^3+3x^2-4x-4=x^3+2x+2$$
$$3x^2-6x-6=0$$
$$x^2-2x-2=0$$



$$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
$$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    so we start off with:
    $$x^3+3x^2-4x-4=x^3+2x+2$$
    which we can re-write as:
    $$3x^2-6x-6=0$$
    so we know that:
    $$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$



    The mistake you made was $36-72$ instead of $36+72$






    share|cite|improve this answer





















    • Thanks but would 108 not be 114 since the initial 6 +-?
      – Doug Fir
      Sep 17 '18 at 15:49






    • 1




      that is not inside the squareroot so we do not include it
      – Henry Lee
      Sep 17 '18 at 15:51










    • mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
      – Henry Lee
      Sep 17 '18 at 15:52
















    2














    so we start off with:
    $$x^3+3x^2-4x-4=x^3+2x+2$$
    which we can re-write as:
    $$3x^2-6x-6=0$$
    so we know that:
    $$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$



    The mistake you made was $36-72$ instead of $36+72$






    share|cite|improve this answer





















    • Thanks but would 108 not be 114 since the initial 6 +-?
      – Doug Fir
      Sep 17 '18 at 15:49






    • 1




      that is not inside the squareroot so we do not include it
      – Henry Lee
      Sep 17 '18 at 15:51










    • mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
      – Henry Lee
      Sep 17 '18 at 15:52














    2












    2








    2






    so we start off with:
    $$x^3+3x^2-4x-4=x^3+2x+2$$
    which we can re-write as:
    $$3x^2-6x-6=0$$
    so we know that:
    $$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$



    The mistake you made was $36-72$ instead of $36+72$






    share|cite|improve this answer












    so we start off with:
    $$x^3+3x^2-4x-4=x^3+2x+2$$
    which we can re-write as:
    $$3x^2-6x-6=0$$
    so we know that:
    $$x=frac{-bpmsqrt{b^2-4ac}}{2a}=frac{6pmsqrt{36-(4)(3)(-6)}}{6}=1pmfrac{sqrt{108}}{6}=1pmsqrt{3}$$



    The mistake you made was $36-72$ instead of $36+72$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 17 '18 at 15:38









    Henry Lee

    1,713218




    1,713218












    • Thanks but would 108 not be 114 since the initial 6 +-?
      – Doug Fir
      Sep 17 '18 at 15:49






    • 1




      that is not inside the squareroot so we do not include it
      – Henry Lee
      Sep 17 '18 at 15:51










    • mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
      – Henry Lee
      Sep 17 '18 at 15:52


















    • Thanks but would 108 not be 114 since the initial 6 +-?
      – Doug Fir
      Sep 17 '18 at 15:49






    • 1




      that is not inside the squareroot so we do not include it
      – Henry Lee
      Sep 17 '18 at 15:51










    • mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
      – Henry Lee
      Sep 17 '18 at 15:52
















    Thanks but would 108 not be 114 since the initial 6 +-?
    – Doug Fir
    Sep 17 '18 at 15:49




    Thanks but would 108 not be 114 since the initial 6 +-?
    – Doug Fir
    Sep 17 '18 at 15:49




    1




    1




    that is not inside the squareroot so we do not include it
    – Henry Lee
    Sep 17 '18 at 15:51




    that is not inside the squareroot so we do not include it
    – Henry Lee
    Sep 17 '18 at 15:51












    mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
    – Henry Lee
    Sep 17 '18 at 15:52




    mathwarehouse.com/dictionary/D-words/… remember where you can apply your commutative and distributive laws
    – Henry Lee
    Sep 17 '18 at 15:52











    1














    $$x^3+3x^2-4x-4=x^3+2x+2$$
    $$3x^2-6x-6=0$$
    $$x^2-2x-2=0$$



    $$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
    $$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$






    share|cite|improve this answer


























      1














      $$x^3+3x^2-4x-4=x^3+2x+2$$
      $$3x^2-6x-6=0$$
      $$x^2-2x-2=0$$



      $$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
      $$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$






      share|cite|improve this answer
























        1












        1








        1






        $$x^3+3x^2-4x-4=x^3+2x+2$$
        $$3x^2-6x-6=0$$
        $$x^2-2x-2=0$$



        $$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
        $$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$






        share|cite|improve this answer












        $$x^3+3x^2-4x-4=x^3+2x+2$$
        $$3x^2-6x-6=0$$
        $$x^2-2x-2=0$$



        $$x=frac{-(-2)pmsqrt{(-2)^2-4(1)(-2)}}{2(1)}$$
        $$x=frac{2pmsqrt{4+8}}{2}=frac{2pm 2sqrt 3}{2}=1pmsqrt 3$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 17 '18 at 15:44









        Deepesh Meena

        4,29921025




        4,29921025






























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