I'm trying to solve a problem that requires me to find the roots of the equation $$-x^4+10x^2-x-20=0$$
By making use of Newton's method, symbolab's equation calculator has found one root to be
$x\approx -2.79129... : x=-\frac{\sqrt{21}+1}{2}$
My question is pretty basic, actually: how did they convert the decimal number on the left to the neat faction on the right? I couldn't get to it the way I've learned to convert decimals. Thanks very much in advance.
On
Almost certainly the polynomial was factored to get $-(x^2+x-5)(x^2-x-4)$ and then the quadratic equation was used to solve for the zeros of each factor.
On
You have accepted another answer, but here is an explanation of how a program or web site could find an expression such as the one given for a floating-point value.
If you have reason to believe that a given floating-point number is a quadratic irrational number, a program can check if that hypothesis is reasonable and, if it is reasonable, give an expression for the value.
The program first finds the continued fraction expansion of the given value. In our case, it is easier to look at $|x|\approx 2.79128784747792$. The first quotient is $\mathrm{int}(x)=2$. To get the later ones, replace $x$ with $\dfrac 1{\mathrm{frac}(x)}$ and take the integer part again. For the given value we get the quotients
$$[2; 1, 3, 1, 3, 1, 3, \ldots]$$
We see that the quotients are repeating. If we keep going, the inherent inaccuracies of the floating-point format will ruin the pattern, but we have good reason to believe that the repetition would be infinite. This repeating pattern in a continued-fraction expansion means the value is a quadratic irrational.
Plain algebra can then be used to find an expression for that value. In this case we see for the repeating part that
$$u=1+\frac 1{3+\frac 1u}$$
Solving that leads to the quadratic equation $3u^2-3u-1=0$, which has the one positive solution $u=\dfrac{3+\sqrt{21}}{6}$, then we find $x$ from
$$x=2+\frac 1u$$
which of course is $x=\dfrac{1+\sqrt{21}}{2}$. This entire process can be automated. Again, the limitations of the floating-point format make this uncertain to get the exact correct value, but using extended-precision arithmetic can make mistakes very few.
On
Bit late; I would emphasize that we can factor the polynomial ourselves, and this one is not difficult. We check that there are no rational roots. After that, we need only try integer constants $a,b,c,d$ with $$ (x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + \cdots $$ Since we are trying to get $x^4 - 10 x^2 + x + 20,$ we see that $c+a = 0.$ $$ (x^2 + ax + b)(x^2 -ax + d) = x^4 + (b+d - a^2)x^2 + a(d-b)x + bd. $$ From $a(d-b)= 1$ we have $a = \pm 1.$ Both work, they just switch $b,d.$ First, $a=1$ gives $$ (x^2 + x + b)(x^2 -x + d) = x^4 + (b+d - 1)x^2 + (d-b)x + bd, $$ so $b+d = -9, d - b = 1.$ Then $2d = -8,$ $d= -4,$ $b=-5,$ and we check $$ (x^2 + x -5)(x^2 -x -4) = x^4 -10 x^2 + x + 20. $$
Repeat: YOU CAN DO THIS
You are making an assumption that they figured the decimal first and converted it the fraction second.
In actuality they did the exact opposite. They used the quadratic formula to get the fraction, Which is accurate and exact. Then they calculated and estimated it to be approximately the decimal on the right.
Converting an irrational decimal to a fraction with the nescessary symbols is impossible for the obvious reason we can not ever have a complete decimal representation of an irrational number.