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





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2920313%2fpolynomial-equation-cannot-solve-for-x%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2920313%2fpolynomial-equation-cannot-solve-for-x%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            How to change which sound is reproduced for terminal bell?

            Image
            3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
            Read more

            Can I use Tabulator js library in my java Spring + Thymeleaf project?

            Image
            0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
            Read more

            Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

            Image
            up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
            Read more