How to solve $x^4+6x^3-14x^2+160=0$

Question

solve $x^4+6x^3-14x^2+160=0$

I used Newton-Raphson and it didn't work for me:

Newton -Raphson : $$x_1=x_0-\frac{f\left(x_0\right)}{f'\left(x_0\right)}$$ $$f(x)=x^4+6x^3-14x^2+160$$ $$f'(x)=4x^3+18x^2-28x$$

with initial guess $x_0 = 1$, I got:

\begin{align}&x_1 = 26.5\\ &x_2 \approx 19.60676\\ &x_3 \approx 14.4647900462\\ &x_4 \approx 10.639431558896779194\\ & \dots \\ &\dots \\ &\dots \\ & x_{10} \approx 39.01424\\ \end{align}

The answers for $f(x)=0$ are $\boxed{x\approx -2.64478\dots ,x\approx -7.48852\dots}$

Can someone help me solve it using this method or any method?

Answer 1

In applying the Newton-Raphson method it is essential to identify a suitable initial value. This is as much an art as a science, but there is one foolproof way to get the method to converge:

• Let $x_\infty$ be a zero of $f(x)$. If $f(x)$ and $(x-x_\infty)f'(x)$ are both one-signed and identically signed between this zero and some other value $x*$, then an initial value between $x_\infty$ and $x*$, or at $x*$ itself, guarantees convergence to $x_\infty$.

Since the sign of $f'(x)$ enters into this theorem, we should seek the critical points of the polynomial function. Fortunately the derivative function given in the question factors into $x$ times a quadratic, from which we find that the critical points are at $0, (-9\pm\sqrt{193})/4$. The function value is found to be positive at $x=0$ and at $x=(-9+\sqrt{193})/4$ but negative at $x=(-9-\sqrt{193})/4$, so we should expect two negative roots that straddle the $x=(-9-\sqrt{193})/4$ critical point. Call the absolutely larger one $A$ and the other one $B$.

For the $A$ root, note that both $f$ and $(x-A)f'$ will be positive when $x<A$, so we may render $x*\to-\infty$ and select any initial value less than $x=(-9-\sqrt{193})/4$ for which the function is positive. Trial with integer values for $x$ reveals $-8$ is suitable, and it gives ready convergence to the $-7.49$ root.

Getting the B root is trickier. One method to do this to "wall off" the known $A$ root by introducing an asymptote in between, say by dividing the polynomial by $x+7$. Using the quotient rule for differentiation we easily verify that

$g(x) := f(x)/(x+7), g'(x)=[f'(x)/(x+7)]-[f(x)/(x+7)^2]$

from which our Newton iteration may be rendered directly in terms of the original polynomial function $f$:

$x_{n+1}=x_n-\dfrac{g(x_n)}{g'(x_n)}=x_n-\dfrac{f(x_n)}{f'(x_n)-\dfrac{f(x)}{x+7}}$ (*)

Then when $x$ lies between $-7$ and $B$, the above scheme will give convergence to $B$ because both $g(x)$ and $(x-B)g'(x)$ will be negative. The appropriate initial value is then one that satisfies $-7<x<0$ and $f(x)<0$, the latter implying $g(x)<0$. Thus for instance $-5$ as an initial value, with the iteration formula (*) given above, gives convergence to the $-2.64$ root.

Answer 2

Since $f(-3)=-47$ and $f(-2)=72$, you can be sure that there is a root in $(-3,-2)$. And, if you apply the Newton-Raphson with $x_0=-2$, you shall get

• $x_1=-2.75$;
• $x_2\approx-2.64637$;
• $x_3\approx-2.64479$;
• $x_4\approx-2.64479$

and so on.

Answer 3

It is really hard to choose a priori the right initial approximation, one that will lead to convergence to a certain root. Proximity is a good criteria, but the overall process is very complex. In the pictures below, the plane is divided into regions, each one coloured according to the root that will be obtained taking a certain point as initial approximation. The second picture is just a zoom of the center of the first one.

As an example, initial approximations in the green areas will produce sequences that convergence to the root $x = -2.6...$

Answer 4

Darboux theorem says that it we start iterating at a point $x_0$ such that $$f(x_0)\times f''(x_0) >0$$ the path to solution will be monotonic and that we shall never overshoot the solution.

For your problem and the largest negative root, this happens for $x_0 < -7.5$.

Using a poor estimate $x_0= -10$, the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & -10.0000 \\ 1 & -8.56250 \\ 2 & -7.78255 \\ 3 & -7.51877 \\ 4 & -7.48889 \\ 5 & -7.48852 \end{array} \right)$$

For the other root $x_0=-3$ satisfies the condition and the iterates are

$$\left( \begin{array}{cc} n & x_n \\ 0 & -3.00000 \\ 1 & -2.65942 \\ 2 & -2.64482 \\ 3 & -2.64479 \end{array} \right)$$

Answer 5

The equation has two real roots and two imaginary ones. The real ones are: $x_{1}=-\sqrt{R}-\sqrt{T+\sqrt{S}}-c$,

$x_{2}=-\sqrt{R}+\sqrt{T+\sqrt{S}}-c$;

the parameter values are:

$R=A^{1/3}+a^{1/3}+b$,

$S=h+g^{1/3}+G^{1/3}-f^{1/3}-F^{1/3}$,

$T=j-k^{1/3}-K^{1/3}$;

further:

$F=\frac{3237158375}{5832}-\frac{166375 \sqrt{19765}}{54}$,

$f=\frac{3237158375}{5832}+\frac{166375 \sqrt{19765}}{54}$,

$G=\frac{ 609113809}{729}-\frac{ 155656 \sqrt{19765}}{27}$,

$g=\frac{ 609113809}{729}+\frac{ 155656 \sqrt{19765}}{27}$,

$h=\frac{101}{4}$,

$K=\frac{19457}{216}-\frac{\sqrt{19765}}{2}$,

$k=\frac{19457}{216}+\frac{\sqrt{19765}}{2}$,

$b=\frac{55}{12}$,

$c=\frac{3}{2}$.