cubic equation:  $$ax^3+bx^2+cx+d=0$$  Cube means 3.  Quadratic equation:  $$ax^2+bx+c=0$$  Quad means 4. But it is a power of two, should it be called bi... Equation.     My question is: Is this a mistake in naming the quadratic equation?
123621
Why is it called a quadratic if it only contains $x^2$, not $x^4$? Should it not be bidratic or didratic etc?
120317
How would I show the following.  For all odd integers a,b, and c if z is a solution of $ax^2+bx+c=0$ the z is not rational.  How would I show this? This is what I did  Let a be an odd integer then for a integer k it has the form $a=2k+1$  then $b=2t+1$ for if c integer.  then $c=2r+1$ if s is an integer then I do  $(2k+1)(x)^2+2t+1(x)+(2r+1)=0$  But I find my self stuck on what to do any hint are appreciated.
35757
A formula for solving a polynomial of degree three, see  link; $ax^3+bx^2+cx+d=0$, is  $$\begin{align} x\quad&amp;=\quad       \sqrt[3]{           \left(               \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}           \right)            +                 \sqrt{               \left(                   \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}               \right) ^2                 +                \left(                   \frac{c}{3a} - \frac{b^2}{9a^2}               \right) ^3            }       }\\       &amp;+\quad       \sqrt[3]{           \left(               \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}           \right)            -                 \sqrt{               \left(                   \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}               \right) ^2                 +                \left(                   \frac{c}{3a} - \frac{b^2}{9a^2}               \right) ^3            }       }       \;-\;\frac{b}{3a} \end{align}$$     Unlike quadratic, cubic, and quartic polynomials, the general quintic   cannot be solved algebraically in terms of a finite number of   additions, subtractions, multiplications, divisions, and root   extractions, as rigorously demonstrated by Abel (Abel's impossibility   theorem) and Galois. However, certain classes of quintic equations can   be solved [...] Source:    At levels of $5^{\text{th}}$ degree polynomials, things are starting to look really serious in my eyes. My question is:     If it is possible to not answer subjectively: When, where and how often do you find polynomials of higher degrees than two in mathematical, pure/applied, research?
29124
Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$  I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case, $(a+b)x + ((a+b)^2 -3ab)y =1.$  I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions?
92362
Question is to check :  For any real number $c$, the polynomial $x^3+x+c$ has exactly one real root .  the way in which i have proceeded is :  let $a$ be one real root for $x^3+x+c$ i.e., we have $a^3+a+c=0$  i have seen that $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a$  But, $a^3+a+c=0$. So, $-a^3-a=c$.  so, $(x-a)(x^2+ax+(a^2+1))=x^3+x-a^3-a=x^3+x+c$  i.e, $(x-a)(x^2+ax+(a^2+1))=x^3+x+c$  Now, determinant for quadratic $x^2+ax+(a^2+1)$ is $a^2-4(a^2+1)=-3a^2-4 &lt;0$ for any real number $a$  Thus, quadratic has no real root and so is the cubic $x^3+x+c$  I would like to know if this justification is sufficient and if this can be generalized.  I mean, can i say $x^3+ax^2+bx+c$ for $a,b,c\in \mathbb{Z}$ have exactly one real root. (Conditions apply)  can this be generalized to any odd degree polynomial (at least for some special cases)
59344
I have 2 equation in terms of $a,b$ and $c$ .  $3a + 10b + 5c =0$ and $4a + 6b + 2c =0$  I need to find a:b:c and answer is $\dfrac a5 = \dfrac {b}{-7} = \dfrac {c}{11}$  I want to know how to get that?  My attempt:  Given equations can be written in form   $$\left ( \begin{matrix}           3 &amp; 10 &amp; 5 \\           4 &amp; 6 &amp;  2 \\           \end{matrix} \right )  \left ( \begin{matrix}           a\\         b\\         c\\         \end{matrix} \right ) =  \left ( \begin{matrix}            0 \\          0\\         \end{matrix} \right ) $$  But here I don't have any idea how to proceed. I can do for 3×3 matrix and I thought it can be done in same way.  Thanks.
108090
Three positive real numbers $a,b,c$ satisfy the equations $a^2+\sqrt3 ab+b^2=25$,  $b^2+c^2=9$  and $a^2+ac+c^2=16$ .Then find the value of $ab+ 2cb+\sqrt3 ac$?  Is there some way to find the desired value without actually finding values of $a,b,c$ or any other smart method to find $a,b,c$
72096
Suppose the quadratic polynomial  $P(x) = ax^2 + bx +c $ has positive coefficients $ a, b, c $ in an arithmetic progression in that order. If $P(x) = 0$ has integer roots $\alpha $ and $\beta $, then $\alpha + \beta + \alpha\beta $ equals?  1) $3$  2) $5$  3) $7$  4) $14$  It's a single choice correct problem. All genuine answers are welcome :)  The question is from a famous Indian Scholarship test 'KVPY' for high school students. The official website doesn't provide any solutions to it that is why I am asking for help.  My go on the question   we know that in a quadratic equation $ax^2 + bx +c$ with roots $\alpha, \beta$ the sum of roots is $\alpha +\beta = \frac{(-b)}{a} $ and product is $\alpha\beta = \frac{c}{a} $  So $\alpha + \beta + \alpha\beta  = \frac{(-b+c)}{a}$ and since $a,b,c$ are in arithmetic progression $\frac{(a+c)}{2} = b$  Something that I tried was also this -  $  a, b, c     $   are in AP (arithmetic progression)  $a-b , 0, c-b$ are in  AP (subtracting by $b$)  $1-\frac{b}{a}, 0, \frac{(c-b)}{a}$ are in AP  (dividing by $a$)  Now I can't proceed further. After a bit of plugging and chugging I still cant get an integer as an answer...
39674
Question :     The coefficients a,b,c of the quadratic equation $ax^2+bx+c=0$ are   determined by throwing a die three times and reading off the value   shown on the uppermost face of each die. i.e, if you throw a 1, 5 and   3 respectively, the equation is $1x^2+5x+3=0$.      Find the probabilities that the quadratic equation you obtain :         has real roots;   has complex roots;   has equal roots.      Thank you for your attention.
74091
Problem: Show that if $(b,c)=1$, then for any integer $a$ we have $(a,bc)=(a,b)(a,c)$. [Hint: Prove that each member of the alleged equation divides the other.]  This is the way I proved it (without the hint): $(a,b)(a,c)=(a(a,b),c(a,b))=(a^2,ab,ac,bc)=(a^2,bc,a(b,c))=(a^2,a,bc)=((a^2,a),bc)=(a,bc)$  But how should I prove it using that hint or at least a similar proof.
90917
I'd like to solve $AX -BX + XC = D$, for the matrix $X$, where all matrices have real entries and $X$ is a rectangular matrix, while $B$ and $C$ are symmetric matrices and $A$ is formed by an outer product matrix (i.e, as $vv^T$ for some real vector $v$) while $D$ is 'not' symmetric. $A,B,C,D$ matrices are fixed while $X$ is the unknown.  How can this equation be solved? Secondly, is there any case, where the solution of this equation has a closed form?
33329
Is there any way to tell whether a quartic equation has double or triple root(s)? $$x^4 + a x^3 + b x^2 + c x + d = 0$$
15167
The terms $a,b,c$ of quadratic equation $ax^{2}+bx+c=0$ are in A.P. and positive. Let this equation have integral root $\alpha,\ \beta$. Then find the value of $\alpha+ \beta + \alpha \cdot \beta$ ?   please point where I'm wrong:  Let common difference be $d$  $\implies \alpha+ \beta + \alpha \cdot \beta=\dfrac{c-b}{a}=\dfrac{d}{a} \implies a|d \ \ \ \ \longrightarrow  \ \ \ \ \              \because (b=a+d$, $c=a+2d)$  Also , $ax^{2}+(a+d)x+(a+2d)=0$.   $\implies$ $\alpha,\ \beta=\dfrac{-(a+d) \pm \sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}}{2a}$.   For this to be integer $\sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}$ must be perfect square.  $\implies$ ${(a+d)^{2}-4\cdot a \cdot (a+2d)}=p^{2}$ for some $p$.  $\implies -3a^{2}+d^{2}-6ad=p^{2}$  $\implies -3a^{2}+a^{2}q^{2}-6a^{2}q=p^{2}$ $\because$ $a|d \implies aq=d$ for some $q$.  $\implies a^{2}(-3+q^{2}-6q)=p^{2}$   $\implies -3+q^{2}-6q\ $  has to be perfect square. By trial $q=7$  But I need to get this without trial, please help.
5923
In my text book, to solve cubic equations, I need to    find by trial &amp; error what $f(a)$ will make the equation 0. The factor will be $(x-a)$ then the other factor will be $Ax^2+Bx+C$ then I can solve using compare coefficient   But for te first part, is there any other way apart from trail &amp; error? Or is there some technique to guess the factor of an equation?   The question I am currently doing happens to be $2x^3+3x+4=9$   UPDATE:  $2x^3+3x+4=9$ should be $2x^3+3x+4=9x^2$   UPDATE 2  So I am going to try substituting   $x=\pm{1}, \pm\frac{1}{2}, \pm{2}, \pm{4}$  When $x=-\frac{1}{2}$, one of the provided answers,   $2(−0.5)^3−9(−0.5)^2+3(−0.5)+4=-18.75\neq 0$   did I make a mistake?
85727
Using the quadratic formula, we have either 0, 1, or 2 solutions. I wonder if we could use it this formula for congruence? Or is there a formula to solve quadratic equation for congruence?  Edit Assume that $ax^2 + bx + c \equiv 0 \pmod{p}$, where $p$ is prime with $(a, p) = 1$, then is there a formula for this equation?    Thanks,
117667
So everyone knows that when $ax^2+bx+c=0$,$$x = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ But why does it equal this? I learned this in maths not 2 weeks ago and it makes no sense to me
95812
I was taught that you could only complete the square of a quadratic if the coefficient on the $x^2$ term is 1.   However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CTS algorithm, you just need to divide the coefficient of the $x$ term by twice the square root of the coefficient of the $x^2$ term.  So, if you have $ax^2 + bx + c$, your perfect square would be $(\sqrt{a}x + \frac{b}{2 \sqrt{a}})^2$  If $a$ is not a perfect square it could get nasty, but then you can just square the whole quadratic and go from there.  For example:   In the equation $5x^2 + 6x + 5 = 0$, we could do:  $25x^2 + 30x + 25 = 0$  $(5x+3)^2 = -16$  $5x+3 = \pm4i$  $x = \pm \frac{4i}{5} - \frac{3}{5}$   My questions are:  -Is this correct?  -Is there ever an advantage to using the quadratic formula?  -Are there quadratics that are unsolvable this way?