Finding the roots of 4096x^3-10496x^2+152576x - 961=0 (1 root and 2 complex)?

Question

I don't know how to find the roots of 4096x^3-10496x^2+152576x - 961=0

I try using wolfram and http://en.wikipedia.org/wiki/Cubic_function. I don't really understand it can someone please explain how it is done?

Answer 1

A cubic equation in the form $y^3+py^2+qy+r=0$ may be reduced to the form $$ x^3+ax+b+0$$ by substituting for $y$ the value $x-\frac{p}{3}$. Here $$ a = \frac{1}{3}(3q-p^2) \quad \text{and} \quad b = \frac{1}{27}(2p^3-9pq+27r)$$ For the solution let $$A = \sqrt[3]{-\frac{b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}$$ and $$ B = \sqrt[3]{-\frac{b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}$$ Then the values of $x$ will be given by $$ x_1 = A+B \\ x_2 = -\frac{A+B}{2}+\frac{A-B}{2} \sqrt{-3} \\ x_3 = - \frac{A+B}{2} - \frac{A-B}{2}\sqrt{-3}$$ Recall that $y=x-\frac{p}{3}$, so you solve for $y_1,y_2$, and $y_3$ by using the corresponding $x$ values.

If $p,q,r$ are real, then let $ d = \dfrac{b^2}{4} + \dfrac{a^3}{27}$.

(i) If $d>0$, then there will be one real root and two conjugate complex roots.

(ii) If $d=0$, then there will be three real roots of which at least two are equal.

(iii) If $d<0$, then there will be three real and unequal roots.

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For your problem, we can say your equation is $$4096y^3-10496y^2+152576y - 961=0 \\ y^3 - \frac{10496}{4096}y^2+\frac{152576}{4096}y - \frac{961}{4096} = 0$$ Therefore, your values of $p,q$, and $r$ are $$p = -\frac{10496}{4096} = -\frac{41}{16} \\ q = \frac{152576}{4096} = \frac{149}{4} \\ r = - \frac{961}{4096}$$ So $$ a = \frac{1}{3}(3q-p^2) = \frac{26927}{768}$$ and $$ b = \frac{1}{27}(2p^3-9pq+27r) \approx 30.33668801 $$

We then define $$A = \sqrt[3]{-\frac{b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}} \approx 3.020887293 $$ and $$B = \sqrt[3]{-\frac{b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}} \approx -3.868752734 $$

Solving for $x_1, x_2,$ and $x_3$: $$ x_1 = A+B = -0.847865441\\ x_2 = -\frac{A+B}{2}+\frac{A-B}{2} \sqrt{-3} = 0.4239327205+5.966603286i \\ x_3 = - \frac{A+B}{2} - \frac{A-B}{2}\sqrt{-3} = 0.4239327205-5.966603286i$$

Therefore, the roots of your equation are $$y_1 = x_1-\frac{p}{3} = 0.0063012257 = \mathbf{0.0063012} \\ y_2 = x_2-\frac{p}{3} = 1.278099387+5.966603286i = \mathbf{1.2781+5.9666i} \\ y_3 = x_3-\frac{p}{3} = 1.278099387-5.966603286i = \mathbf{1.2781-5.9666i}$$

Which fits under case (i) with $d>0$. WolframAlpha's answers to your original expression are listed in bold above, which are indeed equal to the ones we have derived.

Answer 2

The Wikipedia page you refer to shows how to go from a x^3 + b x^2 + c x + d = 0 to t^3 + p t + q = 0 via change of variable (from x to t). Now, the test of the sign of (4 p^3 + 27 q^2) tells how are the roots.