I made a little progress:

BigSams wrote: I made a little progress:

You´re wrong The second equation is not one of the Vietes equations. It´s lacking some terms (there must be 6 terms) The second equation factors to : $(y+t)(z+x)=4 $ Setting $u=z+x$ we get $u(5-u)=4$ so $u=1$ $or 4$. WLOG $u=4$, then the third equation becomes $\frac{4}{xz} + \frac{1}{yt} = -3$ or $4y(1-y) + x(4-x) =-3$ Setting $m=y(1-y)$ and $n=x(4-x)$ we have that $4m+n=-3$ and $mn=-1$(from the fourth equation). From there you can solve it for $m$ and $n$ and consequently for $x$ and $y$ and you´re done

BigSams wrote: y $a_5$ gives $x^4-5x^3+4x^2-3x-1=0$ Factoring gives us the roots which are.... I can't factor it lol. Anyone want to give it a shot?[/hide] You know you can factor it into 2 quadratic terms, so try to make an educated guess what they should be like. I guessed $ (x^2 + ax+1)(x^2+bx-1) $ because the constant and leading terms come out right that way. Multiply that out and equate coefficients of powers of x to find a and b.

BigSams wrote: I made a little progress:

it is $(x^2 - 4 x - 1) (x^2 - x + 1) = 0$

WakeUp wrote: Find all real solutions to the following system of equations: \[\begin{cases} x+y+z+t=5\\xy+yz+zt+tx=4\\xyz+yzt+ztx+txy=3\\xyzt=-1\end{cases}\] Let's define $\sigma_{i}$ as the elementary symmetric functions in one variable . Therefore , it follows that the solutions to the given system of equations are precisely the zeros of the quartic $p(t)=t^4 -5t^3+4t^2-3t-1$ Now we attempt to factorise it , noting the behaviour of the polynomial at points $t= \pm1 $ we can conclude irreducibility over $\mathbb{Q}$ Then based on structural insight suppose we've the factorisation $p(t)=(t^2 -at +b)(t^2 -ct+d)$ in $\mathbb{Z}[t]$ Comparing coefficients we get $$\begin{aligned}bd=-1 \\ a+c=5 \\ b+d+ac=4 \\ ad+bc=3 \end{aligned}$$Using the fact that the coefficients of the quadratics involved in the factorisation are assumed to be $\in \mathbb{Z}$ we get the solution $(a,b,c,d)=(1,1,4,-1)$ which works and we end up with the factorisation $p(t)=(t^2 -t +1)(t^2 -4t-1)$ Then it's easy to verify that the first quadratic in the representation of $p(t)$ is irreducible over $\mathbb{R}$. Hence we can conclude that the given system has no solutions in $\mathbb{R}$ Point out irregularities if any

BigSams wrote: I made a little progress:

I tried the RRT theorem.. So I guess, all the roots here are irrational. There must be another method to solve the problem.

WakeUp wrote: Find all real solutions to the following system of equations: \[\begin{cases} x+y+z+t=5\\xy+yz+zt+tx=4\\xyz+yzt+ztx+txy=3\\xyzt=-1\end{cases}\] $xy+yz+zt+tx$ is not symmetric in $x,y,z,t$. The "solutions" using quartics are false.

Define $a=x+z,b=y+t,c=xz,d=yt$. Then we have \begin{align*} a+b=5\\ ab=4\\ ad+bc=3\\ cd=-1 \end{align*}and the rest is easy.