Hot answers tagged

190 votes
Accepted

Why can a quadratic equation have only 2 roots?

Suppose there are three distinct roots $x,y,z$. One has $$\begin{cases}ax^2+bx+c=0\\ay^2+by+c=0\\az^2+bz+c=0\end{cases}\Rightarrow\begin{cases}a(x^2-y^2)+b(x-y)=0\\a(x^2-z^2)+b(x-z)=0\end{cases}\...
Piquito's user avatar
  • 29.1k
133 votes

Why does the discriminant in the Quadratic Formula reveal the number of real solutions?

Think about it geometrically $-$ then compute. Everyone knows $x^2$ describes a parabola with its apex at $(0,0)$. By adding a parameter $\alpha$, we can move the parabola up and down: $x^2+\alpha$ ...
M. Winter's user avatar
  • 29.8k
72 votes

Why does Wolfram Alpha say the roots of a cubic involve square roots of negative numbers, when all three roots are real?

This is an example of the casus irreducibilis: the formulas for solving cubics need complex numbers when the cubic has three real roots.
lhf's user avatar
  • 216k
71 votes
Accepted

Monic polynomials whose roots are their remaining coefficients

For the quadratic case $x^2+ax+b=0$, by Viète's formulas we have $$-a=a+b\qquad b=ab$$ The first formula implies $b=-2a$. Substituting this into the second equation gives $-2a=-2a^2$ or $a=a^2$, so $a=...
Parcly Taxel's user avatar
67 votes

Function with no roots

$\exp(x)$ has no real roots, and no complex roots either. It is not difficult to find functions that have no roots: for any function $f(x)$, $g(x) = |f(x)|+1$ has no roots.
Wouter's user avatar
  • 7,653
65 votes
Accepted

A "new" general formula for the quadratic equation?

This is a very useful formula for when you want to accurately find the roots of a quadratic equation in which $a$ might be very small using finite precision arithmetic (e.g. on a computer). It's ...
Milo Brandt's user avatar
  • 60.8k
62 votes
Accepted

Why does the discriminant in the Quadratic Formula reveal the number of real solutions?

Because for $a\neq0$ we obtain: $$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right).$$ Now, we see that if $b^2-4ac<0$ then $\...
Michael Rozenberg's user avatar
55 votes

Why can a quadratic equation have only 2 roots?

I think derivation of quadratic formula is not enough.... Yes it is. The derivation is of the form if $ax^2+bx+c=0$, then $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. The derivation is a proof if you pay ...
djechlin's user avatar
  • 5,366
55 votes
Accepted

Is there an easier way to prove the Fundamental Theorem of Algebra for polynomials with real coefficients?

The fundamental theorem of algebra is exactly as difficult for real vs. complex coefficients. The reason is that if $f(x) = f_0 + \dots + f_n x^n$ is any polynomial with complex coefficients, then the ...
Qiaochu Yuan's user avatar
52 votes

Does Newton's method converge for all polynomials?

Here is a plot of the function $$f(x) = 49 x^7+31 x^6-10 x^5-41 x^4+37 x^3-21 x^2-9 x+12$$ in the region of interest. The blue curve is $f$. The orange lines represent the forward orbit of $x_0 = \...
heropup's user avatar
  • 134k
51 votes
Accepted

Function with no roots

The answer to the question as asked is simply "yes", as others have said. I'd like to give a little more context and explain why (for one interpretation of the question) the answer comes close to ...
Gareth McCaughan's user avatar
50 votes
Accepted

Do $3/8$ (37.5%) of Quadratics Have No $x$-Intercepts?

The problem isn't that there's no way to choose a random quadratic. The problem is that there's many ways, and there's no obvious reason to think of any of them as "the" way to pick one. For starters,...
Ben Millwood's user avatar
  • 14.1k
48 votes
Accepted

Why can you find the roots a of polynomial by factoring it?

It's not an assumption. It is a general fact that if $ab = 0$, then either $a = 0$ or $b = 0$. The roots of a polynomial are the numbers for which that polynomial evaluates to $0$, so to find the ...
A. Thomas Yerger's user avatar
47 votes
Accepted

Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$

There is no need to use Vieta's formulas. Let $$f(z)=z^4-2z^3+z^2+z-7=(z-\alpha)(z-\beta)(z-\gamma)(z-\delta).$$ Then, since $(i-a)(-i-a)=-i^2+a^2=1+a^2$, it follows that $$(\alpha^2+1)(\beta ^2+1)(\...
Robert Z's user avatar
  • 146k
45 votes
Accepted

Cubic formula gives the wrong result (triple checked)

The problem occurs when you are taking a cube root: in the Wolfram version, when you compute $$T = \sqrt[3]{R - \sqrt{D}} = \sqrt[3]{-\frac{35}{27}-\sqrt{\frac{50}{27}}}$$ and in corresponding places ...
Misha Lavrov's user avatar
45 votes

Do $3/8$ (37.5%) of Quadratics Have No $x$-Intercepts?

It's helpful to think of each quadratic polynomial as corresponding to a point in "coefficient space", an abstract 3D space whose coordinates are $(a,b,c)$. The question "What fraction of quadratic ...
Michael Seifert's user avatar
43 votes
Accepted

Influence of small constant term on roots of polynomial

Consider the polynomial, $$p_1(x)=(x-1)(x-2)...(x-19)(x-20).$$ After expanding it, you get, \begin{array}{|r|r|} \hline \textrm{Exponent} & \textrm{Coefficients of $p(x)$}\\\hline 0 & ...
Fixed Point's user avatar
  • 7,899
41 votes
Accepted

Relation between the roots of a function

$$f(x)=(x-r_1)(x-r_2)\cdots(x-r_n)$$ Using logarithm, $$\ln (f(x))=\ln (x-r_1)+\ln(x-r_2)+\cdots +\ln(x-r_n)$$ Taking derivative, $$\frac{f'(x)}{f(x)}= \frac{1}{(x-r_1)}+\frac{1}{(x-r_2)} +\...
Jaideep Khare's user avatar
39 votes
Accepted

Why does Wolfram Alpha say the roots of a cubic involve square roots of negative numbers, when all three roots are real?

When the discriminant of the equation is negative, there are $3$ real roots. It's precisely in this case that Cardano's formulæ do not work, because the radicands are negative. It is even ...
Bernard's user avatar
  • 175k
39 votes
Accepted

Why can't the quadratic formula be simplified to $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm(b-2)\sqrt{ac}}{2a}$?

Because $\sqrt{b^2 - 4ac}\neq 2b\sqrt{ac}$. Say for example that $c = 0$ and $b\neq 0$. Then you have \begin{align*} \sqrt{b^2 - 4ac} &= \sqrt{b^2}\\ &= \left|b\right|\\ &\neq 0, \end{...
Stahl's user avatar
  • 23.2k
39 votes

What does this converge to and why?

When we ask "What does this expression converge to?" we usually mean "If you cut off this infinite expression after a certain point, you get a number. As you cut off the expression later and later, ...
Misha Lavrov's user avatar
39 votes
Accepted

A General Way of Finding the Zeroes of a Polynomial

Now, my doubt is : can the zeroes of any polynomial be found using such forms (as given above)? You are asking a very deep question - one of the main problems of mathematics in the 18th and 19th ...
Randy Marsh's user avatar
  • 2,827
36 votes
Accepted

Can we permute the coefficients of a polynomial so that it has NO real roots?

Yes: put the $n+1$ largest coefficients on the even powers of $x$, and the $n$ smallest coefficients on the odd powers of $x$. Clearly the polynomial will have no nonnegative roots regardless of the ...
Greg Martin's user avatar
  • 77.6k
36 votes
Accepted

Must imaginary roots come in conjugate pairs?

The book's answer is nonsense. Your calculation gives a polynomial with roots at $0,\,4,\,i$ as desired and is a very direct way to find the minimal polynomial that will have a root at all of those ...
Milo Brandt's user avatar
  • 60.8k
36 votes

A "new" general formula for the quadratic equation?

What's interesting here isn't your result, it's your technique. Ask a typical algebra teacher, "What's important about the quadratic formula?", they'll probably say, "The fact that it lets you find ...
pokep's user avatar
  • 501
34 votes
Accepted

Finding the roots $x^4-4x^3-x^2-8x+4=0$ (contest math)

Divide $x^4-4x^3-x^2-8x+4=0$ with $x^2$ in order to get the following $x^2-4x-1-\frac{8}{x}+\frac{4}{x^2}=(x+\frac{2}{x})^2-4(x+\frac{2}{x})-5=(x+\frac{2}{x}-5)(x+\frac{2}{x}+1)$. Finally, result is $(...
alans's user avatar
  • 6,455
34 votes
Accepted

Is there an intuitive way of visualising complex roots?

If your question is related to quadratic equations then here is the solution. Let's take your example. Now, do the following in the case of a parabolic curve not intersecting the $x$ axis. Mirror the ...
zoli's user avatar
  • 20.4k
33 votes

Relation between the roots of a function

While Jaideep Khare's computation with the logarithm gives a good quick way to memorize or derive the result, there is an issue: The logarithms of negative numbers are not defined if one works over ...
Joonas Ilmavirta's user avatar
31 votes

If a polynomial $P$ has only real roots, so does $P'-2xP$

Let $g(x)=P(x)e^{-x^2}$. Then $$g'(x)=P'(x)\cdot e^{-x^2}+P(x)\cdot (-2x)e^{-x^2}= (P'(x)-2xP(x))e^{-x^2}. $$ We may consider $g$ to be extended to the infinities per $g(\pm\infty)=0$ because the ...
Hagen von Eitzen's user avatar

Only top scored, non community-wiki answers of a minimum length are eligible