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

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)?

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What do you mean by "2 and $\frac{1}{2}$ degree"? A polynomial is an expression of the form $a_nx^n+...+a_0$, so the degree is always a non-negative integer. – Dennis Gulko Jul 20 '11 at 16:23
$x^{3/2}=x\sqrt{x}$ – Américo Tavares Jul 20 '11 at 16:23
You might see en.wikipedia.org/wiki/Polynomial. Polynomials are defined with integer exponents. Please give an example of what you mean by "a polynomial with 2 and 1/2 degree." Also, one does not solve a polynomial, one solves an equation. – Ross Millikan Jul 20 '11 at 16:23
What does "a polynomial with 2 and 1/2 degree" mean? For example $x^{5/2}-x=28$ or $x^2-x^{1/2}=14$? – Henry Jul 20 '11 at 16:24
@Victor: If you are complaining about why people have to memorize the quadratic formula when it is of such limited applicability, the answer is that in fact second degree equations come up, in several guises, very often (much like linear equations do) and so the quadratic formula turns out to be a very useful thing to have "at your fingertips" (much like memorizing multiplication tables). – Arturo Magidin Jul 20 '11 at 17:24

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!

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very clear and through!thanks very much! – Victor Jul 20 '11 at 17:23

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.

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You can use Andre's method above and rewrite this as an honest polynomial in some $y=x^{m/n}$. Then, there's an explicit formula iff the degree is less than 5. – Aaron Mazel-Gee Jul 20 '11 at 17:29
The existence of an explicit formula is somewhat more complicated. There is a formula for the roots of the polynomial $x^8 - 1$ namely $$e^{\frac{\pi i n}{4}}$$ for $n = 0, 1, \cdots, 7$. – Jay Jul 20 '11 at 17:41
The result of Abel is that there is no general formula for solving the quintic (I won't define general formula!). The result of Galois, some time later, is that in addition there are specific quintics with integer coefficients which have no root in any extension by radicals. (And he did much more.) Certain limited types of polynomial equations in any degree are solvable by radicals. – André Nicolas Jul 20 '11 at 17:56
@Jay: Sorry, I was being imprecise. Of course certain polynomials have explicit solutions. Since the original question was about general formulas, that's what I meant: for an equation $\sum a_i x^{\mu_i/\nu_i}=0$, there's a general formula for $y=x^{m/n}$ as above exactly when $lcm(\nu_i)\cdot \min\{\mu_i\} \leq 4$. – Aaron Mazel-Gee Jul 20 '11 at 18:33