275. Some interesting algebraic equations and systems.
by Virgil Nicula, May 8, 2011, 1:34 PM
I. Exercises. Solve for 
the following equations : ![$\left\{\begin{array}{cc}
\mathrm{Eq.\ 1\ :} & 4x-x^2=3 \sqrt{4-3\sqrt{10-3x}}\ .\\\\
\mathrm{Eq.\ 2\ :} & \sqrt[3]{1-x}+\sqrt[3]{1+x}=\frac {x^2+2}{\sqrt{x^2+1}}\ .\end{array}\right\|$](//latex.artofproblemsolving.com/f/2/9/f29ff77321d24b38ce072bcd919e8efe67c8a541.png)
Proof of Eq. 1. Observe that
. Denote
, where ![$x\in [2,4]$](//latex.artofproblemsolving.com/b/b/a/bba04c025ce8f837af1bcd7a213a65805c41401f.png)
. Observe that 
, where the functions 
, 
are strict increasing on![$[2,4]$](//latex.artofproblemsolving.com/7/5/c/75ce239185eb37bd138cdb9440bc3f662b5b25c7.png)
. Hence and the function 
is strict increasing. Since 
obtain that our equation 
has at most one zero , i.e. 
.
Proof of Eq. 2. Denote the solution-set
of this equation. Observe that 
. Thus, can suppose w.l.o.g. that 
. From the A.M.-G.M. inequality
obtain that
for any with equality iff 
. If 
, then ![$\sqrt[3]{1-x}+\sqrt[3]{1+x}<2$](//latex.artofproblemsolving.com/9/b/d/9bdedf97d37dde7566a7a2963c4da440e2e05c7e.png)
. Hence our equation
hasn't zeroes in
. For 
we have 
![$\sqrt[3]{x^2+1}\ge$](//latex.artofproblemsolving.com/9/4/f/94fcf6ffdff14b106b0d370fb67d5f0440ea5365.png)
![$ \sqrt[3]{x+1}\ge $](//latex.artofproblemsolving.com/e/6/2/e62a1846429054696df0f3b2dc6a32a1845d7c4b.png)
![$\sqrt[3]{x+1}+\sqrt[3]{1-x}$](//latex.artofproblemsolving.com/3/7/c/37cc4b0bd3742f26598badd1d57855f169c48126.png)
, absurd.
In conclusion, our equation has only unique zero .
II. Exercises. Solve for the following equations : 
.
Proof of Eq. 3. Since
obtain that 
and 
- solution-set. Our equation begins ![$3^x\cdot\left[\left(\frac 43\right)^x-1\right]=2-\sqrt 3$](//latex.artofproblemsolving.com/1/6/d/16d545fec7e860a66bc8ac357933fb3c5f3d04d0.png)
,
where
, 
and 
. Since the functions 
and 
are positive and strict increasing on 
obtain that and
their product
is positive and strict increasing on . In conclusion, our equation 
has only the zero 
.
Proof of Eq. 4. Consider the functions
, 
(straight line) , where . Observe that our equation becomes 
, where
and 
. Since the function is convex obtain that for any 
we have 
and for any ![$x\not \in [2,3]$](//latex.artofproblemsolving.com/9/7/d/97d76116ae1b5cfba0eb7a73c811010e35b983e0.png)
we have 
.
In conclusion, our equation has only two zeroes ,
and 
.
Proof of Eq. 5. Prove easily that![$\{0,1\}\subset S\subset [0,1]$](//latex.artofproblemsolving.com/c/b/e/cbe960532e8c59388f9cf6e738c9d0db21e66fe5.png)
, where is the solution-set.
For apply the A.M.-G.M.-inequality : ![$\left\{\begin{array}{c}
2^x+2^{\sqrt {1-x^2}}=2\cdot 2^{x-1}+2^{\sqrt {1-x^2}}>3\cdot \sqrt[3]{2^{2(x-1)+\sqrt{1-x^2}}}\equiv 3\cdot u(x)\\\\
2^x+2^{\sqrt {1-x^2}}=2^x+2\cdot 2^{\sqrt {1-x^2}-1}>3\cdot \sqrt[3]{2^{x+2\left(\sqrt{1-x^2}-1\right)}}\equiv 3\cdot v(x)\end{array}\right\|$](//latex.artofproblemsolving.com/9/8/c/98cb163332d845c9f4b64794e4eb0c4bb3ba34c6.png)
. Observe that
for any 
.
III. Exercises. Solve for the following equations : 
.
Proof of Eq. 6. Apply the substitutions
and 
, i.e. 
and 
. Our equation becomes 
and obtain the following system :
. Thus, 
from where results 
and
. From 
, 
obtain 
and 
. From the first equations of the system 
obtain that 
.
Thus, the initial equation
becomes 
, i.e. 
. Since 
and
, and obtain that 
, i.e. 
. Observe that 
and
. Denote 
and 
. From 
, 
and 
obtain that
, i.e. 
. Thus, 
, i.e. or 
. In conclusion, our equation has only zero .
I denoted
the real numbers 
, 
have same sign.
Proof of Eq. 7. I"ll study the monotony of the function
, with the parameter . Denote 
, 
and 
.
Observe that , the function 
is strict decreasing on ![$(0,1]$](//latex.artofproblemsolving.com/1/0/1/101678300d1aa11009468da8419aaf2c57b795bd.png)
and strict increasing on 
and the function 
is strict
increasing. Observe that
![$b^{\phi (x)}\left[b^{\psi (x)}+b^{-\psi (x)}\right]$](//latex.artofproblemsolving.com/3/d/0/3d09cc9896c982051becfc19b849868508b17d56.png)
. Hence ![$f(x)=b^{\phi (x)}\cdot \phi \left[b^{\psi (x)}\right]$](//latex.artofproblemsolving.com/a/9/8/a98b2331d1489e1434050719c5b75f4e4df6180b.png)
, where
. If denote 
, where , then ![$f(x)=h[\phi (x)]\cdot \phi \{h[\psi (x)]\}$](//latex.artofproblemsolving.com/8/e/2/8e296a0fcf0b52a4816eb82601f2789f56646875.png)
, where the functions 
and 
are positive.
Since for![$x\in (0,1]$](//latex.artofproblemsolving.com/8/8/5/885ef0336d15614e1f1fcac9cf1aa48751c547e1.png)
these functions are strict decreasing and positive obtain that and its product is strict decreasing. Since for 
these functions
are strict increasing and positive obtain that and its product is strict increasing. Therefore, 
for any , i.e. 
.
In conclusion : if
our equation hasn't real zeroes ; if 
, then 
; if 
, then our equation has only two real zeroes.
Proof of Eq. 8. Denote
, where 
. Prove easily that for any 
we have 
. If and 
,
then and
. Thus can suppose w.l.o.g. that 
. Observe that 
. Our equation becomes 
, i.e. 
, where 
and 
. Observe that
and the functions 
and 
are strict increasing. Therefore,
and 
. Hence 
and 
.for any , what
means
and 
. In conclusion, the solutions of this equation are only 
and 
.
![\[\underline{\overbrace{\alpha\ <\ x_1\ \le\ x_2\ <\ \beta}}\]](//latex.artofproblemsolving.com/f/f/d/ffdc69c6f36ef229a6a18cc4c45a07ae447b553e.png)
Lemma. Let 
be an equation with real coefficients and 
. Given are two real numbers 
.
Prove that the roots
, 
of are really and 
, where 
.
Example.
and 
![$\lambda\in (3,4]$](//latex.artofproblemsolving.com/d/4/2/d42503f9515ca7b41b921c73231784c275e64c62.png)
.
IV Exercise. Given positive reals
, find all real solutions 
of the system 
.
Proof.
. Our system becomes
. Observe that 
. Suppose w.l.o.g. 
.
Hence
. If , then 
, what is false. In conclusion, , i.e. 
.
Remark. It is the proposed problem nr.13.3/pag.115 (Cap. II - Functii reale de vaiabila reala) from my book
"Analiza Matematica pentru clasa a XI - a. Teorie, exercitii si probleme" - Editura TEORA Educational Bucuresti, 1999.
V. Exercise.
such that 
Proof 1. Observe that
. Appear two cases:
Proof 2. Suppose
. Thus our system becomes 
, false.
VI. Exercise. Solve the following system of equation
.
Proof. Exists
so that 
, i.e. 
at least two from
are equally. Suppose w.l.o.g. that 
. In conclusion, 
.
PP7. Solve the system of equations
.
Proof. By substitutions
our system becomes 
, where
. Therefore, 
.
Remark. Let
and 
so that 
. Sum 
is minimum iff 
. In this case 
and is named the Fermat point w.r.t. . Since the minimum point is unique we can solve with geometry this system. I"ll use the relations 
,
where![$S=[ABC]$](//latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
- the area of . Indeed, prove easily that 
and 
. In conclusion,
. From the upper chain 
obtain that 
.
the following equations : ![$\left\{\begin{array}{cc}
\mathrm{Eq.\ 1\ :} & 4x-x^2=3 \sqrt{4-3\sqrt{10-3x}}\ .\\\\
\mathrm{Eq.\ 2\ :} & \sqrt[3]{1-x}+\sqrt[3]{1+x}=\frac {x^2+2}{\sqrt{x^2+1}}\ .\end{array}\right\|$](http://latex.artofproblemsolving.com/f/2/9/f29ff77321d24b38ce072bcd919e8efe67c8a541.png)
Proof of Eq. 1. Observe that
. Denote
, where ![$x\in [2,4]$](http://latex.artofproblemsolving.com/b/b/a/bba04c025ce8f837af1bcd7a213a65805c41401f.png)
. Observe that 
, where the functions 
, 
are strict increasing on
![$[2,4]$](http://latex.artofproblemsolving.com/7/5/c/75ce239185eb37bd138cdb9440bc3f662b5b25c7.png)
. Hence and the function 
is strict increasing. Since 
obtain that our equation 
has at most one zero , i.e. 
.Proof of Eq. 2. Denote the solution-set
of this equation. Observe that 
. Thus, can suppose w.l.o.g. that 
. From the A.M.-G.M. inequalityobtain that
for any with equality iff 
. If 
, then ![$\sqrt[3]{1-x}+\sqrt[3]{1+x}<2$](http://latex.artofproblemsolving.com/9/b/d/9bdedf97d37dde7566a7a2963c4da440e2e05c7e.png)
. Hence our equationhasn't zeroes in
. For 
we have 
![$\sqrt[3]{x^2+1}\ge$](http://latex.artofproblemsolving.com/9/4/f/94fcf6ffdff14b106b0d370fb67d5f0440ea5365.png)
![$ \sqrt[3]{x+1}\ge $](http://latex.artofproblemsolving.com/e/6/2/e62a1846429054696df0f3b2dc6a32a1845d7c4b.png)
![$\sqrt[3]{x+1}+\sqrt[3]{1-x}$](http://latex.artofproblemsolving.com/3/7/c/37cc4b0bd3742f26598badd1d57855f169c48126.png)
, absurd.In conclusion, our equation has only unique zero
.II. Exercises. Solve for
the following equations : 
.Proof of Eq. 3. Since
obtain that 
and 
- solution-set. Our equation begins ![$3^x\cdot\left[\left(\frac 43\right)^x-1\right]=2-\sqrt 3$](http://latex.artofproblemsolving.com/1/6/d/16d545fec7e860a66bc8ac357933fb3c5f3d04d0.png)
,where
, 
and 
. Since the functions 
and 
are positive and strict increasing on 
obtain that andtheir product
is positive and strict increasing on . In conclusion, our equation 
has only the zero 
.Proof of Eq. 4. Consider the functions
, 
(straight line) , where . Observe that our equation becomes 
, where
and 
. Since the function is convex obtain that for any 
we have 
and for any ![$x\not \in [2,3]$](http://latex.artofproblemsolving.com/9/7/d/97d76116ae1b5cfba0eb7a73c811010e35b983e0.png)
we have 
.In conclusion, our equation has only two zeroes ,
and 
.Proof of Eq. 5. Prove easily that
![$\{0,1\}\subset S\subset [0,1]$](http://latex.artofproblemsolving.com/c/b/e/cbe960532e8c59388f9cf6e738c9d0db21e66fe5.png)
, where is the solution-set.For
apply the A.M.-G.M.-inequality : ![$\left\{\begin{array}{c}
2^x+2^{\sqrt {1-x^2}}=2\cdot 2^{x-1}+2^{\sqrt {1-x^2}}>3\cdot \sqrt[3]{2^{2(x-1)+\sqrt{1-x^2}}}\equiv 3\cdot u(x)\\\\
2^x+2^{\sqrt {1-x^2}}=2^x+2\cdot 2^{\sqrt {1-x^2}-1}>3\cdot \sqrt[3]{2^{x+2\left(\sqrt{1-x^2}-1\right)}}\equiv 3\cdot v(x)\end{array}\right\|$](http://latex.artofproblemsolving.com/9/8/c/98cb163332d845c9f4b64794e4eb0c4bb3ba34c6.png)
. Observe that
for any 
.III. Exercises. Solve for
the following equations : 
.Proof of Eq. 6. Apply the substitutions
and 
, i.e. 
and 
. Our equation becomes 
and obtain the following system :
. Thus, 
from where results 
and
. From 
, 
obtain 
and 
. From the first equations of the system 
obtain that 
.Thus, the initial equation
becomes 
, i.e. 
. Since 
and
, and obtain that 
, i.e. 
. Observe that 
and
. Denote 
and 
. From 
, 
and 
obtain that
, i.e. 
. Thus, 
, i.e. or 
. In conclusion, our equation has only zero .I denoted
the real numbers 
, 
have same sign.Proof of Eq. 7. I"ll study the monotony of the function
, with the parameter . Denote 
, 
and 
.Observe that
, the function 
is strict decreasing on ![$(0,1]$](http://latex.artofproblemsolving.com/1/0/1/101678300d1aa11009468da8419aaf2c57b795bd.png)
and strict increasing on 
and the function 
is strictincreasing. Observe that
![$b^{\phi (x)}\left[b^{\psi (x)}+b^{-\psi (x)}\right]$](http://latex.artofproblemsolving.com/3/d/0/3d09cc9896c982051becfc19b849868508b17d56.png)
. Hence ![$f(x)=b^{\phi (x)}\cdot \phi \left[b^{\psi (x)}\right]$](http://latex.artofproblemsolving.com/a/9/8/a98b2331d1489e1434050719c5b75f4e4df6180b.png)
, where . If denote 
, where , then ![$f(x)=h[\phi (x)]\cdot \phi \{h[\psi (x)]\}$](http://latex.artofproblemsolving.com/8/e/2/8e296a0fcf0b52a4816eb82601f2789f56646875.png)
, where the functions 
and 
are positive.Since for
![$x\in (0,1]$](http://latex.artofproblemsolving.com/8/8/5/885ef0336d15614e1f1fcac9cf1aa48751c547e1.png)
these functions are strict decreasing and positive obtain that and its product is strict decreasing. Since for 
these functionsare strict increasing and positive obtain that and its product
is strict increasing. Therefore, 
for any , i.e. 
.In conclusion : if
our equation hasn't real zeroes ; if 
, then 
; if 
, then our equation has only two real zeroes.Proof of Eq. 8. Denote
, where 
. Prove easily that for any 
we have 
. If and 
,then and
. Thus can suppose w.l.o.g. that 
. Observe that 
. Our equation becomes 
, i.e. 
, where 
and 
. Observe that
and the functions 
and 
are strict increasing. Therefore,
and 
. Hence 
and 
.for any , whatmeans
and 
. In conclusion, the solutions of this equation are only 
and 
. ![\[\underline{\overbrace{\alpha\ <\ x_1\ \le\ x_2\ <\ \beta}}\]](http://latex.artofproblemsolving.com/f/f/d/ffdc69c6f36ef229a6a18cc4c45a07ae447b553e.png)
Lemma. Let 
be an equation with real coefficients and 
. Given are two real numbers 
.Prove that the roots
, 
of are really and 
, where 
.Example.
and 
![$\lambda\in (3,4]$](http://latex.artofproblemsolving.com/d/4/2/d42503f9515ca7b41b921c73231784c275e64c62.png)
.IV Exercise. Given positive reals
, find all real solutions 
of the system 
.Proof.
. Our system becomes
. Observe that 
. Suppose w.l.o.g. 
.Hence
. If , then 
, what is false. In conclusion, , i.e. 
.Remark. It is the proposed problem nr.13.3/pag.115 (Cap. II - Functii reale de vaiabila reala) from my book
"Analiza Matematica pentru clasa a XI - a. Teorie, exercitii si probleme" - Editura TEORA Educational Bucuresti, 1999.
V. Exercise.
such that 
Proof 1. Observe that
. Appear two cases:
Proof 2. Suppose
. Thus our system becomes 
, false. VI. Exercise. Solve the following system of equation
.Proof. Exists
so that 
, i.e. 
at least two from
are equally. Suppose w.l.o.g. that 
. In conclusion, 
.PP7. Solve the system of equations
.Proof. By substitutions
our system becomes 
, where
. Therefore, 
.Remark. Let
and 
so that 
. Sum 
is minimum iff 
. In this case 
and
is named the Fermat point w.r.t. . Since the minimum point is unique we can solve with geometry this system. I"ll use the relations 
,where
![$S=[ABC]$](http://latex.artofproblemsolving.com/b/3/a/b3ae3d445111e4dd28be75922309d3270079368c.png)
- the area of . Indeed, prove easily that 
and 
. In conclusion,
. From the upper chain 
obtain that 
.This post has been edited 126 times. Last edited by Virgil Nicula, Nov 22, 2015, 7:52 AM
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