• Sep 22nd 2009, 09:29 AM
dhiab
Solve in R :
$\sqrt {x - 1} + \sqrt {2 - x} = x^2 - 3x + 3$
• Sep 22nd 2009, 09:32 AM
stapel
Square both sides. You'll end up with a radical as the middle term on the right-hand side.

Isolate this radical, and then square again.

Then solve the resulting polynomial.

• Sep 22nd 2009, 09:43 AM
dhiab
Quote:

Originally Posted by stapel
Square both sides. You'll end up with a radical as the middle term on the right-hand side.

Isolate this radical, and then square again.

Then solve the resulting polynomial.

Hello :I'have
$\begin{array}{l}
\forall x \in \Re :x^2 - 3x + 3 > 0 \\
because:\Delta = - 3 < 0 \\
then: \\
x - 1 + 2 - x + 2\sqrt {\left( {x - 1} \right)\left( {2 - x} \right)} = \left( {x^2 - 3x + 3} \right)^2 \\
2\sqrt {\left( {x - 1} \right)\left( {2 - x} \right)} = \left( {x^2 - 3x + 3} \right)^2 - 1??????? \\
\end{array}$
• Sep 22nd 2009, 10:45 AM
red_dog
$\sqrt{x-1}+\sqrt{2-x}=-(x-1)(2-x)+1$

Let $a=\sqrt{x-1}, \ b=\sqrt{2-x}$

Then we have the system of equations:

$\left\{\begin{array}{ll}a+b=-a^2b^2+1\\a^2+b^2=1\end{array}\right.$

Let $S=a+b, \ P=ab$. Then

$\left\{\begin{array}{ll}S+P^2=1\\S^2-2P=1\end{array}\right.$

From the first equation $S=1-P^2$

Replace in the second:

$P(P^3-2P-2)=0$

If $P=0$ then $S=1$.

Then $\left\{\begin{array}{ll}a+b=1\\ab=0\end{array}\rig ht.$

We get $\left\{\begin{array}{ll}a=1\\b=0\end{array}\right. \Rightarrow x=2$

$\left\{\begin{array}{ll}a=0\\b=1\end{array}\right. \Rightarrow x=1$

Now we have to see what happens if $P^3-2P^2-2=0$

But this equation has no rational roots and the discussion is more complicated.
• Sep 22nd 2009, 07:15 PM
pacman
from red_dog, p^3 - 2p^2 - 0 = 0, by iteration; p > 2.359.

the graph confirms it
• Sep 22nd 2009, 10:38 PM
dhiab
Quote:

Originally Posted by red_dog
$\sqrt{x-1}+\sqrt{2-x}=-(x-1)(2-x)+1$

Let $a=\sqrt{x-1}, \ b=\sqrt{2-x}$

Then we have the system of equations:

$\left\{\begin{array}{ll}a+b=-a^2b^2+1\\a^2+b^2=1\end{array}\right.$

Let $S=a+b, \ P=ab$. Then

$\left\{\begin{array}{ll}S+P^2=1\\S^2-2P=1\end{array}\right.$

From the first equation $S=1-P^2$

Replace in the second:

$P(P^3-2P-2)=0$

If $P=0$ then $S=1$.

Then $\left\{\begin{array}{ll}a+b=1\\ab=0\end{array}\rig ht.$

We get $\left\{\begin{array}{ll}a=1\\b=0\end{array}\right. \Rightarrow x=2$

$\left\{\begin{array}{ll}a=0\\b=1\end{array}\right. \Rightarrow x=1$

Now we have to see what happens if $P^3-2P^2-2=0$

But this equation has no rational roots and the discussion is more complicated.

HELLO :
This equation $p^3 - 2p^2 - 2 = 0
$
has not real roots because I'have $0 \le p \le 1$
and if

$p \in \left[ {0,1} \right]:p^3 - 2p^2 - 2 < 0$
There is the details :

$1 \le x \le 2.....because:\left( {x - 1 \ge 0} \right) \wedge \left( {2 - x \ge 0} \right)
$

Then :
$\begin{array}{l}
\\
1 \le x \le 2 \Leftrightarrow 0 \le x - 1 \le 1 \Leftrightarrow 0 \le \sqrt {x - 1} \le 1......(1) \\
1 \le x \le 2 \Leftrightarrow - 1 \ge - x \ge - 2 \Leftrightarrow 0 \le 2 - x \le 1 \Leftrightarrow 0 \le \sqrt {2 - x} \le 1.....(2) \\
\end{array}$

$but:a = \sqrt {x - 1} ...b = \sqrt {x - 1}$
by (1),(2) :

$0 \le p \le 1
$

THANKS