Why people have to find quadratic formula,isn't that the formula cannot solve a polynomial with 2 and 1/2 degree?

Question

Why people have to find quadratic formula,isn't that the formula cannot solve a polynomial with 2 and 1/2 degree? and just curious, how many roots does a polynomial with 2 and 1/2 degree have and how to solve them all(by formula)?

Answer 1

A polynomial $P(x)$ in one indeterminate with coefficients in a ring $A$ is an expression of the form $$a_nx^n+a_1x^{n-1}+ \cdots + a_0$$ where $n$ is an integer $\ge 0$ and the coefficients $a_0$, $a_1$, $\dots$, $a_n$ are elements of the ring $A$.

Familiar examples are polynomials with integer coefficients (the ring $A$ is then the ring $\mathbb{Z}$ of integers), polynomials with rational coefficients (the ring $A$ is then the ring $\mathbb{Q}$ of rational numbers), and polynomials with real coefficients.

There are much more "exotic" examples, of great theoretical and practical importance. For example, polynomials with coefficients in the set $\{0,1\}$, with addition and multiplication defined modulo $2$, are important in coding theory. The arithmetic here is very simple, $0+0=0\;$, $0+1=1+0=1\;$, $1+1=0\;$, $0\cdot 0=0\;$, $0\cdot 1=1\cdot 0=0\;$, $1\cdot 1=1$.

In addition to polynomials with one indeterminate, we can also consider polynomials in two indeterminates, such as $3x^2y^3-7xy+17$, three indeterminates, and so on.

The expression $$6x^{5/2} -4x^{3/2} -x+17$$ is therefore technically not a polynomial. However, in this example, let $y=x^{1/2}$. Then we can rewrite the above expression as $$6y^5 -4y^3-y^2 +17$$ which is definitely a polynomial in $y$. We ordinarily say that $6x^{5/2} -4x^{3/2} -x+17$ is not a polynomial "in" $x$, but is a polynomial in $x^{1/2}$.

Not that this helps if we want to find roots, since solving even polynomial equations "by formula" is hopeless by degree $5$.

The Quadratic Formula can be adapted to solve some equations that strictly speaking are not polynomial equations. For example, look at the equation $$x-x^{1/2}-1=0.$$

Make the substitution $y=x^{1/2}$. We obtain the equation $$y^2-y-1=0$$ which has the solutions $$y=\frac{1\pm\sqrt{5}}{2}.$$

So now we know the possible values of $x^{1/2}$, and we can square them to get the possible values of $x$.

But please remember that when the plain word "polynomial" is used, the exponents (powers) used are non-negative integers.

About your question on the number of solutions of a polynomial equation of "degree" $5/2$, the first thing I would say is that it is not a polynomial equation. In certain cases, you would be able to make a substitution, like our earlier $y=x^{1/2}$, and turn the equation into a polynomial equation in $y$. But depending on the values of other exponents, the number of solutions could be large.

And if you look for example at the equation $x^{5/2}-x^{\pi/2}+1=0$, no substitution will turn this into a polynomial equation.

I hope this was not overly elaborate and technical!

Answer 2

As Ross points out in his comment polynomials are defined to have exponents that are positive integers. If you find the roots of a polynomial $p(x)$ using this definition than, in certain cases, one can find the roots by algebraic means. A polynomial of degree $n$ has $n + 1$ different coefficients, the constant term together with the coefficient of each power of $x$. For each degree of the polynomial there is a limited amount of variability. If one allows fractional exponents this is no longer the case. Consider the following list of "fractional degree polynomials."

• $x^{\frac{5}{2}} + 1 = 0$
• $x^{\frac{5}{2}} + 2x^{\frac{1}{2}} + 1 = 0$
• $x^{\frac{5}{2}} + 2x^{\frac{1}{2}} + 3x^{\frac{1}{3}} + 1 = 0$
• $x^{\frac{5}{2}} + 2x^{\frac{1}{2}} + 3x^{\frac{1}{3}} + 4x^{\frac{1}{4}} + 1 = 0$
• $x^{\frac{5}{2}} + 2x^{\frac{1}{2}} + 3x^{\frac{1}{3}} + 4x^{\frac{1}{4}} + 5x^{\frac{1}{5}} + 1 = 0$
• and so forth.

All of these might be considered a "polynomial" of degree $\frac{5}{2}$. There is no finite limit to the number of terms such a "polynomial" could have. This means there is no finite limit to the number of coefficients such a "polynomial" has. It seems to me that it is unlikely there is a single algebraic formula that could find the roots of all of these "polynomials." The Rolling Stones have a song about this: You can't always get what you want.