Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can anyone help me solve this?

Approximate the solutions to two decimal places:


Help will be appreciated.

share|cite|improve this question
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
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
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
up vote 0 down vote accepted

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


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

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

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

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

share|cite|improve this answer

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:


share|cite|improve this answer

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.