$x^4 +6x^3 + 11x^2 +7x +1= (1-x)^{0.5}$
I was wondering if anyone could solve this equation and give the answer is fractions. Also please mention the way you solved it. 1 solution is obviously $0$ and the other I got using graphs are in decimals and not precise. It is this second solution in $\Bbb R$ that I require. If it isn't mathematically possible to a get a fractional solution please let me know . Thank you.
This is not the full solution.
Square both sides:
$$\left(x^{4}+6x^{3}+11x^{2}+7x+1\right)^{2}=\color{blue}{\left(\sqrt{1-x}\right)^{2}}$$
$$\color{red}{x^{8}}+\color{red}{x^{7}}\left(6+6\right)+\color{red}{x^{6}}\left(11+36+11\right)+\color{red}{x^{5}}\left(7+66+66+7\right)+\color{red}{x^{4}}\left(1+42+121+42+1\right)+\color{red}{x^{3}}\left(6+77+77+6\right)+\color{red}{x^{2}}\left(11+49+11\right)+\color{red}{x}\left(7+7\right)+\left(1\right)=\color{blue}{1-x}$$
Also for calculating the LHS you could use multinomial theorem.
Equivalently $$x^8+12x^7+58x^6+146x^5+207x^4+166x^3+71x^2+14x+1=1-x$$
Or $$x^8+12x^7+58x^6+146x^5+207x^4+166x^3+71x^2+15x=0$$ Assume rational roots of this equations are in the form $\frac{p}{q}$ where $p,q∈ℤ$ and $q≠0$, also assume this fraction is in the simplest form (GCD of $p,q$ is $1$), using rational root theorem implies $p$ must divide $15$ and $q$ must divide $1$, so the whole fractions with these assumptions are:
$$\pm1 , \pm3 ,\pm5$$
Checking them implies $\color{green} {\boxed {x=-3}}$ is one of the solutions.
Now you can divide the equation by $x+3$ to get $$x\left(x^6+9x^5+31x^4+53x^3+48x^2+22x+5\right)$$
clearly $\color{green} {\boxed {x=0}}$ is the other solution.
Also for solving $x^6+9x^5+31x^4+53x^3+48x^2+22x+5$ I suggest to use Newton's method, then you can get all the solutions.
For more numericals algorithms (but more difficult ways) you can use , Bisection method , Muller's method,Secant method,false position method
The equation$$x^6+9x^5+31x^4+53x^3+48x^2+22x+5=0$$ has only two real roots $(-3.40)$ and $(-2.26)$.
If you start Newton method with a rational value, it will generate rational values as approximations.
Let us start the first one with $x_0=-\frac 7 2$; the method will generate the sequence $$\left\{-\frac{7}{2},-\frac{17761}{5200},-\frac{760677732007010815845121}{22395520514 5491367398240}\right\}$$ For the second one, start with $x_0=-\frac 9 4$ to get $$\left\{-\frac{9}{4},-\frac{3163}{1400},-\frac{20328497875010225431}{8998263788043807 760}\right\}$$ If you compute very precisely the solutions and make them later rational, you would have $$-\frac{108915225749466541474917137561912526763616184782665}{32074761139291841554652427819810193110590856685177}$$ $$-\frac{113549291667849330530332249667376629095991878162727}{50261780121389507187318172078184361726047088691641}$$ which are correct for $100$ figures.
