Solve in R :
$\displaystyle \sqrt {x - 1} + \sqrt {2 - x} = x^2 - 3x + 3$
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.
Remember to check your solutions in the original radical equation!
If you get stuck, please reply showing your steps and reasoning so far. Thank you!
Hello :I'have
$\displaystyle \begin{array}{l}
\forall x \in \Re :x^2 - 3x + 3 > 0 \\
because:\Delta = - 3 < 0 \\
then: \\
x - 1 + 2 - x + 2\sqrt {\left( {x - 1} \right)\left( {2 - x} \right)} = \left( {x^2 - 3x + 3} \right)^2 \\
2\sqrt {\left( {x - 1} \right)\left( {2 - x} \right)} = \left( {x^2 - 3x + 3} \right)^2 - 1??????? \\
\end{array}$
$\displaystyle \sqrt{x-1}+\sqrt{2-x}=-(x-1)(2-x)+1$
Let $\displaystyle a=\sqrt{x-1}, \ b=\sqrt{2-x}$
Then we have the system of equations:
$\displaystyle \left\{\begin{array}{ll}a+b=-a^2b^2+1\\a^2+b^2=1\end{array}\right.$
Let $\displaystyle S=a+b, \ P=ab$. Then
$\displaystyle \left\{\begin{array}{ll}S+P^2=1\\S^2-2P=1\end{array}\right.$
From the first equation $\displaystyle S=1-P^2$
Replace in the second:
$\displaystyle P(P^3-2P-2)=0$
If $\displaystyle P=0$ then $\displaystyle S=1$.
Then $\displaystyle \left\{\begin{array}{ll}a+b=1\\ab=0\end{array}\rig ht.$
We get $\displaystyle \left\{\begin{array}{ll}a=1\\b=0\end{array}\right. \Rightarrow x=2$
$\displaystyle \left\{\begin{array}{ll}a=0\\b=1\end{array}\right. \Rightarrow x=1$
Now we have to see what happens if $\displaystyle P^3-2P^2-2=0$
But this equation has no rational roots and the discussion is more complicated.
HELLO :
This equation $\displaystyle p^3 - 2p^2 - 2 = 0
$ has not real roots because I'have $\displaystyle 0 \le p \le 1$
and if
$\displaystyle p \in \left[ {0,1} \right]^3 - 2p^2 - 2 < 0$
There is the details :
$\displaystyle 1 \le x \le 2.....because:\left( {x - 1 \ge 0} \right) \wedge \left( {2 - x \ge 0} \right)
$
Then :
$\displaystyle \begin{array}{l}
\\
1 \le x \le 2 \Leftrightarrow 0 \le x - 1 \le 1 \Leftrightarrow 0 \le \sqrt {x - 1} \le 1......(1) \\
1 \le x \le 2 \Leftrightarrow - 1 \ge - x \ge - 2 \Leftrightarrow 0 \le 2 - x \le 1 \Leftrightarrow 0 \le \sqrt {2 - x} \le 1.....(2) \\
\end{array}$
$\displaystyle but:a = \sqrt {x - 1} ...b = \sqrt {x - 1} $
by (1),(2) :
$\displaystyle 0 \le p \le 1
$
THANKS