401. Some nice problems with polynomials/equations II.
by Virgil Nicula, Nov 9, 2014, 11:05 AM
PP0. Let 
be two real numbers so that 
, 
and 
. Express 
as a quadratic expression in 
with rational coefficients and vice versa.
Proof. Let
be the roots of 
. Prove easily that 
(with calculus), where 
.
Observe that
, 
(symmetrical expression in and ) . I"ll find 
so that
. Thus, from the relation 
obtain that 
. Using 
obtain 
. Hence
and 
. Eliminate 
between 
and 
. I"ll eliminate the variable 
between 
and 
. Thus,
=============================================================================================
================================================================================================================
======================================================================================================================
======================================================================================================================
.
These last two equations from
must have at least one common root, i.e. 
. (see here).
So the equations what appear in have at least a common root for 
iff 
.
the equations 
have the common root 
and from the relation 
obtain that 
. In conclusion, 
.
have the common root 
and from obtain that 
. In conclusion, 
.
Remark.
have at least the common root 
. Thus, 
.
PP1. Find the range of the function
, where 
.
Proof. Range of
is the set of all 
for which 
has at least a zero 
, i.e. exists at least 
so that 
, i.e. 
. Let the equation ,
where
. Since 
for any 
(
is odd) can suppose w.l.o.g. 
. Thus, 
. The equation , where 
becomes
, i.e. 
, where 
. Thus,
. Hence 
, i.e. ![$\lambda\in (0,1]\cup [2,\infty )$](//latex.artofproblemsolving.com/1/3/9/139cdadca20c078cf76b528d8c32108bc078cffc.png)
.
Prove easily that
, i.e. 
. Thus obtain at most a possible zero 
. From the relation obtain 
. In conclusion, the range of the function on 
is 
. Hence the range of the function on 
is 
.
PP2. Solve the equation
, where 
.
Proof 1. Observe that the given equation is equivalently with
. Appear two cases:
.
.
Proof 2. I"ll use the substitution
, where 
. Thus, 
and the equation becomes 
.
Observe that
is not a solution and so the equation is equivalent to 
and 
and
or 
. Hence the result is 
.
PP3. Find all values of
such that the equation 
has only real roots, where 
Proof. Since
, i.e. 
can write 
,
where denoted
. Thus, has all real roots iff the equation 
has all real roots and greater or equally than 
. Define the following
equivalence relation over
, i.e. 
. Prove easily that 
we have 
and 
. In conclusion, the equation 
has all real roots and greater or equally
than
, i.e. the equation has all real roots iff 
.
==================================================================================================================
Denote 
. Thus, 
![$x^3\cdot\left[\left(x^3+\frac {1}{x^3}\right)+3\cdot\left(x^2+\frac {1}{x^2}\right)+6\cdot\left(x+\frac 1x\right)+7\right]=$](//latex.artofproblemsolving.com/d/c/2/dc2c0aca888579068a55692efe3b774ff691fb07.png)
![$x^3\cdot\left[\left(t^3-3t\right)+3\cdot\left(t^2-2\right)+6t+7\right]=$](//latex.artofproblemsolving.com/b/b/f/bbf10977d6c1c576579726f86f871bb0471d98ff.png)
![$[x(t+1)]^3=(xt+x)^3=$](//latex.artofproblemsolving.com/5/c/3/5c33f0d531585b864b0fb3ee560f095a6694d837.png)
![$\left[x\left(x+\frac 1x\right)+x\right]^3=\left(x^2+x+1\right)^3$](//latex.artofproblemsolving.com/3/c/4/3c46ddf5ec88b8703154c92a1bb3b1fe38e4caf0.png)
.
Remark. Observe that
. Indeed, 
.
PP4. Let![$p\in \mathbb R[X]$](//latex.artofproblemsolving.com/0/8/6/0867b67ffffa686c4829333dcb91db13bcd0ea60.png)
be a polynomial so that 
. Prove that ![$(\exists )\ \{m,n\}\subset R[X]$](//latex.artofproblemsolving.com/9/1/3/913dfea720302558815712e927a92740cc76ceb3.png)
so that 
.
Proof. Suppose w.l.o.g. that
has the dominant coefficient 
and 
for any , i.e. 
. Is obviously that the degree of is even,
i.e.
and for any 
we have 
. Hence 
![$\prod_{k=1}^s\left[\left(x-x_k\right)-i\cdot y_k\right]\cdot\prod_1^s\left[\left(x-x_k\right)+i\cdot y_k\right]=$](//latex.artofproblemsolving.com/8/8/6/886e1173fa95fcf9ed0f25a00f571280ac5d1bd7.png)
, where 
for any 
and ![$u(x)=\prod_{k=1}^s\left[\left(x-x_k\right)-i\cdot y_k\right]\in\mathbb C[X]$](//latex.artofproblemsolving.com/3/6/a/36ad211a9330beaa1821decfd21680bc80920672.png)
. Since exists two polynomials ![$\{m,n\}\subset \mathbb R[X]$](//latex.artofproblemsolving.com/4/c/d/4cd6df054123fc822d45f9d9bc0f1fa3f3a0168f.png)
so that
obtain that 
.
Application. If![$p\in\mathbb R[X]$](//latex.artofproblemsolving.com/f/c/4/fc49894b407296a4230454bc165a4ec8d9a65ce0.png)
such that 
have 
, then 
matrix 
have 
.
Proof. Indeed, from upper property get that exist![$\{m.n\}\subset\mathbb R[X]$](//latex.artofproblemsolving.com/1/1/8/118cdbd919bd754648118a6757e229dd2b6924b6.png)
so that . Hence and 
. Since the matrices 
, 
are permutably, obtain
that![$p(A)=[m(A)+i\cdot n(A)]\cdot [m(A)-i\cdot n(A)]$](//latex.artofproblemsolving.com/0/3/0/030d39ca2dfdefcc2adf73fb1e07c5bc61aa0f34.png)
. Denote 
, where 
. Observe that 
and 
.
In conclusion,
and 
, i.e. .
PP5. Find real number so that the equation 
has only three real distinct zeroes.
Graphical method. Construct easily the graph of the function
. A variable line 
with the equation 
pass through the fixed point 
which belongs to the graph
of the function , i.e. 
is an independent zero of the equation 
. By moving around 
get the points 
and 
which
define the required lines
and 
with the slopes 
. Generally, if denote the solution 
and the number 
of the zeroes for the equation , then 
![$\mathrm{S}_p=\{0\}\cup\left\{\begin{array}{ccc}
\emptyset & \mathrm{if} & p<-1\ \vee\ p\ge 1\\\\
(-\infty , 2]^* & \mathrm{if} & p=-1\\\\
\left\{\frac {4}{1-p}\right\} & \mathrm{if} & -1<p<-\frac 14\\\\
\left\{\frac {16}{5},8\right\} & \mathrm{if} & p=-\frac 14\\\\
\left\{\frac {4}{1-p},\frac {6}{1+p},\frac {10}{1-p}\right\} & \mathrm{if} & -\frac 14<p<\frac 15\\\\
\left\{5,\frac {25}{2}\right\} & \mathrm{if} & p=\frac 15\\\\
\left\{\frac {10}{1-p}\right\} & \mathrm{if} & \frac 15<p<1\\\\
\end{array}\right\|$](//latex.artofproblemsolving.com/f/8/4/f84a65b409995f1e47cb67266140a8870e56dca3.png)
and 
.
Remarks.
Denote the point 
and 
. Observe that ![$\left\{\begin{array}{ccccc}
d_p\cap [DB]\ : & \iff & \left|\begin{array}{c}
y=px+2\\\
y=x-2\end{array}\right| & \iff & x=\frac {4}{1-p}\ ,\ p\in \left[-1,\frac 15\right]\\\\
d_p\cap [BC]\ : & \iff & \left|\begin{array}{c}
y=px+2\\\
y=-x+8\end{array}\right| & \iff & x=\frac {6}{1+p}\ ,\ p\in \left[-\frac 14,\frac 15\right]\\\\
d_p\cap [CE\ : & \iff & \left|\begin{array}{c}
y=px+2\\\
y=x-8\end{array}\right| & \iff & x=\frac {10}{1-p}\ ,\ p\in \left[-\frac 14,1\right)\end{array}\right\|$](//latex.artofproblemsolving.com/5/6/b/56bace7bdd36a3aca74dd57568eab3d4f08a8933.png)
.
The equation , where and
has a maximum number of zeroes if and only if 
.
and 
hasn't zeroes, i.e. the range of 
, where 
is 
.
PP6. Let
. Find 
and a polynomial with integer coefficients with roots 
.
Proof 1 (level IX). I"ll use the Viete's relations
and 
. Now I"ll find 
. Analogously 
. In conclusion, the required equation is 
, i.e. 
.
Proof 2 (level XI).
. Therefore,
Since 
and 
, btain that 
.
Let
. Eliminate between 
and 
. Thus, 
. Observe that we found again 
, i.e. .
be two real numbers so that 
, 
and 
. Express 
as a quadratic expression in 
with rational coefficients and vice versa.Proof. Let
be the roots of 
. Prove easily that 
(with calculus), where 
.Observe that
, 
(symmetrical expression in and ) . I"ll find 
so that
. Thus, from the relation 
obtain that 
. Using 
obtain 
. Hence
and 
. Eliminate 
between 
and 
. I"ll eliminate the variable 
between 
and 
. Thus,=============================================================================================
================================================================================================================
======================================================================================================================
======================================================================================================================
.These last two equations from
must have at least one common root, i.e. 
. (see here).So the equations what appear in
have at least a common root for 
iff 
.
the equations 
have the common root 
and from the relation 
obtain that 
. In conclusion, 
.
have the common root 
and from obtain that 
. In conclusion, 
.Remark.
have at least the common root 
. Thus, 
.PP1. Find the range of the function
, where 
.Proof. Range of
is the set of all 
for which 
has at least a zero 
, i.e. exists at least 
so that 
, i.e. 
. Let the equation ,where
. Since 
for any 
(
is odd) can suppose w.l.o.g. 
. Thus, 
. The equation , where 
becomes
, i.e. 
, where 
. Thus,
. Hence 
, i.e. ![$\lambda\in (0,1]\cup [2,\infty )$](http://latex.artofproblemsolving.com/1/3/9/139cdadca20c078cf76b528d8c32108bc078cffc.png)
.Prove easily that
, i.e. 
. Thus obtain at most a possible zero 
. From the relation obtain 
. In conclusion, the range of the function on 
is 
. Hence the range of the function on 
is 
.PP2. Solve the equation
, where 
.Proof 1. Observe that the given equation is equivalently with
. Appear two cases:
.
.Proof 2. I"ll use the substitution
, where 
. Thus, 
and the equation becomes 
.Observe that
is not a solution and so the equation is equivalent to 
and 
and
or 
. Hence the result is 
.PP3. Find all values of
such that the equation 
has only real roots, where 
Proof. Since
, i.e. 
can write 
,where denoted
. Thus, has all real roots iff the equation 
has all real roots and greater or equally than 
. Define the followingequivalence relation over
, i.e. 
. Prove easily that 
we have 
and 
. In conclusion, the equation 
has all real roots and greater or equallythan
, i.e. the equation has all real roots iff 
.==================================================================================================================
Denote 
. Thus, 
![$x^3\cdot\left[\left(x^3+\frac {1}{x^3}\right)+3\cdot\left(x^2+\frac {1}{x^2}\right)+6\cdot\left(x+\frac 1x\right)+7\right]=$](http://latex.artofproblemsolving.com/d/c/2/dc2c0aca888579068a55692efe3b774ff691fb07.png)
![$x^3\cdot\left[\left(t^3-3t\right)+3\cdot\left(t^2-2\right)+6t+7\right]=$](http://latex.artofproblemsolving.com/b/b/f/bbf10977d6c1c576579726f86f871bb0471d98ff.png)
![$[x(t+1)]^3=(xt+x)^3=$](http://latex.artofproblemsolving.com/5/c/3/5c33f0d531585b864b0fb3ee560f095a6694d837.png)
![$\left[x\left(x+\frac 1x\right)+x\right]^3=\left(x^2+x+1\right)^3$](http://latex.artofproblemsolving.com/3/c/4/3c46ddf5ec88b8703154c92a1bb3b1fe38e4caf0.png)
.Remark. Observe that
. Indeed, 
.PP4. Let
![$p\in \mathbb R[X]$](http://latex.artofproblemsolving.com/0/8/6/0867b67ffffa686c4829333dcb91db13bcd0ea60.png)
be a polynomial so that 
. Prove that ![$(\exists )\ \{m,n\}\subset R[X]$](http://latex.artofproblemsolving.com/9/1/3/913dfea720302558815712e927a92740cc76ceb3.png)
so that 
.Proof. Suppose w.l.o.g. that
has the dominant coefficient 
and 
for any , i.e. 
. Is obviously that the degree of is even,i.e.
and for any 
we have 
. Hence 
![$\prod_{k=1}^s\left[\left(x-x_k\right)-i\cdot y_k\right]\cdot\prod_1^s\left[\left(x-x_k\right)+i\cdot y_k\right]=$](http://latex.artofproblemsolving.com/8/8/6/886e1173fa95fcf9ed0f25a00f571280ac5d1bd7.png)
, where 
for any 
and ![$u(x)=\prod_{k=1}^s\left[\left(x-x_k\right)-i\cdot y_k\right]\in\mathbb C[X]$](http://latex.artofproblemsolving.com/3/6/a/36ad211a9330beaa1821decfd21680bc80920672.png)
. Since exists two polynomials ![$\{m,n\}\subset \mathbb R[X]$](http://latex.artofproblemsolving.com/4/c/d/4cd6df054123fc822d45f9d9bc0f1fa3f3a0168f.png)
so that
obtain that 
.Application. If
![$p\in\mathbb R[X]$](http://latex.artofproblemsolving.com/f/c/4/fc49894b407296a4230454bc165a4ec8d9a65ce0.png)
such that 
have 
, then 
matrix 
have 
.Proof. Indeed, from upper property get that exist
![$\{m.n\}\subset\mathbb R[X]$](http://latex.artofproblemsolving.com/1/1/8/118cdbd919bd754648118a6757e229dd2b6924b6.png)
so that . Hence and 
. Since the matrices 
, 
are permutably, obtainthat
![$p(A)=[m(A)+i\cdot n(A)]\cdot [m(A)-i\cdot n(A)]$](http://latex.artofproblemsolving.com/0/3/0/030d39ca2dfdefcc2adf73fb1e07c5bc61aa0f34.png)
. Denote 
, where 
. Observe that 
and 
.In conclusion,
and 
, i.e. .PP5. Find real number
so that the equation 
has only three real distinct zeroes.Graphical method. Construct easily the graph of the function
. A variable line 
with the equation 
pass through the fixed point 
which belongs to the graph
of the function , i.e. 
is an independent zero of the equation 
. By moving around 
get the points 
and 
whichdefine the required lines
and 
with the slopes 
. Generally, if denote the solution 
and the number 
of the zeroes for the equation , then 
![$\mathrm{S}_p=\{0\}\cup\left\{\begin{array}{ccc}
\emptyset & \mathrm{if} & p<-1\ \vee\ p\ge 1\\\\
(-\infty , 2]^* & \mathrm{if} & p=-1\\\\
\left\{\frac {4}{1-p}\right\} & \mathrm{if} & -1<p<-\frac 14\\\\
\left\{\frac {16}{5},8\right\} & \mathrm{if} & p=-\frac 14\\\\
\left\{\frac {4}{1-p},\frac {6}{1+p},\frac {10}{1-p}\right\} & \mathrm{if} & -\frac 14<p<\frac 15\\\\
\left\{5,\frac {25}{2}\right\} & \mathrm{if} & p=\frac 15\\\\
\left\{\frac {10}{1-p}\right\} & \mathrm{if} & \frac 15<p<1\\\\
\end{array}\right\|$](http://latex.artofproblemsolving.com/f/8/4/f84a65b409995f1e47cb67266140a8870e56dca3.png)
and 
.Remarks.
Denote the point 
and 
. Observe that ![$\left\{\begin{array}{ccccc}
d_p\cap [DB]\ : & \iff & \left|\begin{array}{c}
y=px+2\\\
y=x-2\end{array}\right| & \iff & x=\frac {4}{1-p}\ ,\ p\in \left[-1,\frac 15\right]\\\\
d_p\cap [BC]\ : & \iff & \left|\begin{array}{c}
y=px+2\\\
y=-x+8\end{array}\right| & \iff & x=\frac {6}{1+p}\ ,\ p\in \left[-\frac 14,\frac 15\right]\\\\
d_p\cap [CE\ : & \iff & \left|\begin{array}{c}
y=px+2\\\
y=x-8\end{array}\right| & \iff & x=\frac {10}{1-p}\ ,\ p\in \left[-\frac 14,1\right)\end{array}\right\|$](http://latex.artofproblemsolving.com/5/6/b/56bace7bdd36a3aca74dd57568eab3d4f08a8933.png)
. The equation , where and
has a maximum number of zeroes if and only if 
. 
and 
hasn't zeroes, i.e. the range of 
, where 
is 
.PP6. Let
. Find 
and a polynomial with integer coefficients with roots 
.Proof 1 (level IX). I"ll use the Viete's relations
and 
. Now I"ll find 
. Analogously 
. In conclusion, the required equation is 
, i.e. 
.Proof 2 (level XI).
. Therefore,
Since 
and 
, btain that 
.Let
. Eliminate between 
and 
. Thus, 
. Observe that we found again 
, i.e. .This post has been edited 106 times. Last edited by Virgil Nicula, Jan 20, 2016, 2:33 PM
August 2018
January 2018