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Care este valoarea parametrului rational m, daca ecuatia
\( x^4 - 7x^3 + (13 + m)x^2 - (3 + 4m)x + m = 0 \)
admite solutia \( x_1 = 2 + \sqrt 3 \) si solutiile \( x_3 \) si \( x_4 \) verifica relatia \( x_3 = 2x_4 \).

Se folosesc relatiile lui \( Viete \) folosind faptul ca f admite si radacina \( x_2=2-\sqrt{3} \).

Am incercat cu relatiile lui Viete
\(
\[
\begin{array}{l}
x_1 + x_2 + x_3 + x_4 = \frac{{ - b}}{a} = 7 \Rightarrow 2 + \sqrt 3 + x_2 + 3x_4 = 7 \\
x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4 = 13 + m \\
\end{array}
\]
\)

In a 2-a relatie am inlocuit pe \( x_1 \) cu \( 2+\sqrt3 \) si am facut calculele si am ajuns la rezultatul
\(
\[
(2 + \sqrt 3 )(5 - \sqrt 3 ) + x_4 (3x_2 + 2x_4 ) = 13 + m
\]
\)
si de aici m-am blocat.

\( x_2=2-\sqrt{3} \) deoarece m e rational

Adriany wrote: Care este valoarea parametrului \( m\in\mathbb Q \) , daca ecuatia \( x^4 - 7x^3 + (13 + m)x^2 - (3 + 4m)x + m = 0 \)

admite radacina \( x_1 = 2 + \sqrt 3 \) si radacinile \( x_3 \) si \( x_4 \) verifica relatia \( x_3 = 2x_4\ \ \vee\ \ x_4 = 2x_3 \) .

Dem. Deoarece coeficientii ecuatiei date sunt rationali rezulta ca \( x_3=2-\sqrt 3 \) .

\( \odot\ \ \left\|\ \begin{array}{c}
x_1=2+\sqrt 3\\\\\\\\
x_2=2-\sqrt 3\end{array}\ \ \right\|\ \Longleftrightarrow\ \left\|\ \begin{array}{c}
S=4\\\\\\\\
P=1\end{array}\ \right\|\ \Longleftrightarrow\ \underline{\overline{\left\|\ x^2-4x+1=0\ \begin{array}{cc}
\nearrow & x_1\\\\\\\\
\searrow & x_2\end{array}\ \right\|}} \) .

\( \odot\ \ x_3=2x_4\ \ \vee\ \ x_4=2x_3\ \Longleftrightarrow\ x_3-2x_4=0\ \ \vee\ \ x_4-2x_3=0\ \Longleftrightarrow \)

\( \left(x_3-2x_4\right)\left(x_4-2x_3\right)=0\ \Longleftrightarrow\ 5P=2S_2\ \Longleftrightarrow\ 2S^2=9P\ \Longleftrightarrow \)

exista \( n\in\mathbb Q \) astfel incat \( \left\|\ \begin{array}{c}
S=n\\\\\\\\\
P=\frac {2n^2}{9}\end{array}\ \right\|\ \Longleftrightarrow\ \underline{\overline{\left\|\ x^2-nx+\frac {2n^2}{9}=0\ \begin{array}{cc}
\nearrow & x_3\\\\\\\\
\searrow & x_4\end{array}\ \right\|}} \) .

Deci \( X^4 - 7X^3 + (13 + m)X^2 - (3 + 4m)X + m \equiv\left(X^2-4X+1\right)\left(X^2-nX+\frac {2n^2}{9}\right)\ \Longleftrightarrow \)

\( \left\|\ \begin{array}{ccc}
-n-4=-7 & \Longrightarrow & n=3\\\\\\\\
\frac {2n^2}{9}+4n+1=13+m & \Longrightarrow & m=2\\\\\\\\
-\frac {8n^2}{9}-n=-3-4m & \Longrightarrow & \mathrm{DA}\\\\\\\\
\frac {2n^2}{9}=m & \Longrightarrow & \mathrm{DA}\end{array}\ \right\|\ . \) In concluzie, \( \underline{\overline{\left\|\ m=2\ \right\|}} \) si ecuatia devine

\( \underline{\overline{\left\|\ x^4-7x^3+15x^2-11x+2=\left(x^2-4x+1\right)\left(x^2-3x+2\right)=0\ \begin{array}{c}
\nearrow\\\\\\\\
\searrow\end{array}\ \begin{array}{ccc}
x_1 & = & 2+\sqrt 3\\\\\\\\
x_2 & = & 2-\sqrt 3\\\\\\\\
x_3& = & 1\\\\\\\\
x_4 & = & 2\end{array}\ \right\|}}\ . \)