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Can anyone help me solve this?

Approximate the solutions to two decimal places:

$$x^3-8x-3=0$$

Help will be appreciated.

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Do you mean $x^3-8x-3 = 0$? –  Sam DeHority Jan 29 '13 at 0:49
    
f(x)=((x^(3)-8x-3)) –  Little Jon Jan 29 '13 at 0:51
    
yes that is what I mean good sir. –  Little Jon Jan 29 '13 at 0:52
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This seems like a question in numerical analysis. Do you want to know possible methods or just the answer? –  Tunococ Jan 29 '13 at 0:53
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For approximate solutions to 0s of polynomials I would suggest type the expression x^3 -8x -3 = 0 into Wolfram|Alpha to get an answer. If you're looking for methods of solution then let me know. –  Sam DeHority Jan 29 '13 at 0:54

3 Answers 3

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If you are feeling ambitious and curious you can also find the solutions using Cardano's method to come up with the roots of $x^3-8x-3=0$ with the result as follows

$$\sqrt[3]{\frac{-19i\sqrt{15}}{18}+\frac{3}{2}}+\sqrt[3]{\frac{19i\sqrt{15}}{18}+\frac{3}{2}}$$

Then take the cube roots of the above to obtain the following result

$$\left(\frac{3}{2}+\frac{i\sqrt{15}}{6}\right)+\left(\frac{3}{2}-\frac{i\sqrt{15}}{6}\right) = 3$$

$$\left(\frac{-3-\sqrt{5}}{4}\right)+\left(\frac{9i\sqrt{3}-i\sqrt{15}}{12}\right)+\left(\frac{-3-\sqrt{5}}{4}\right)+\left(\frac{-9i\sqrt{3}+i\sqrt{15}}{12}\right)=\frac{-3-\sqrt{5}}{2} $$

$$\left(\frac{-3+\sqrt{5}}{4}\right)+\left(\frac{9i\sqrt{3}+i\sqrt{15}}{12}\right)+\left(\frac{-3+\sqrt{5}}{4}\right)+\left(\frac{-9i\sqrt{3}-i\sqrt{15}}{12}\right)=\frac{-3+\sqrt{5}}{2} $$

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There is a closed form solution for cubics, but that is not what the question asks. However, in this case, you'll find that there is a root at $x=3$. Using synthetic division, we find that the quotient of the cubic divided by $x-3$ is $x^2 + 3 x+1$, which has roots at $x=\frac{-3 \pm \sqrt{5}}{2}$. You may verify this by a quick inspection of the graph of the cubic:

Cubic

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$$x^3-8x-3=0$$ $$x^3-3x^2+3x^2-9x+x-3=0$$ $$x^2(x-3)+3x(x-3)+1(x-3)=0$$ $$(x^2+3x+1)(x-3)=0$$ $$x-3=0,x_1=3$$ $$x^2+3x+1=0,x_2=\frac{-3+\sqrt{5}}{2},x_3=\frac{-3-\sqrt{5}}{2}$$

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The fact that $3$ is a root could have been detected with the rational roots test. –  Austin Mohr Jan 29 '13 at 1:12
    
Then I divide by $x-3$ and got quadratic –  Adi Dani Jan 29 '13 at 1:14

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