جواب کاردرکلاس صفحه 13 درس 1 ریاضی یازدهم تجربی (هندسۀ تحلیلی و جبر)

\(\begin{array}{l}\left. \begin{array}{l}S = - 1/5\\P = - 7\end{array} \right\} \Rightarrow {x^2} - Sx + P = {x^2} + 1/5x - 7 = 0 \Rightarrow \Delta = {\left( {1/5} \right)^2} - 4\left( 1 \right)\left( { - 7} \right) = 30/25\\x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 1/5 \pm 5/5}}{2} \Rightarrow \left\{ \begin{array}{l}{x_1} = \frac{{ - 1/5 - 5/5}}{2} = - 3/5\\{x_2} = \frac{{ - 1/5 + 5/5}}{2} = 2\end{array} \right.\end{array}\)

الف

\(\begin{array}{l}\alpha \left( {\frac{{11}}{2} - \alpha } \right) = 6 \Rightarrow - {\alpha ^2} + \frac{{11}}{2}\alpha - 6 = 0 \Rightarrow \Delta = {\left( {\frac{{11}}{2}} \right)^2} - 4\left( { - 1} \right)\left( { - 6} \right) = 6/25\\\alpha = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{ - 5/5 \pm \sqrt {6/25} }}{{ - 2}} \Rightarrow \left\{ \begin{array}{l}{\alpha _1} = \frac{{ - 5/5 - 2/5}}{{ - 2}} = 4\\{\alpha _2} = \frac{{ - 5/5 + 2/5}}{{ - 2}} = - 1/5\end{array} \right.\\ \Rightarrow \alpha = 4 \Rightarrow \beta = 1/5\end{array}\)

ب

\(\begin{array}{l}\left. \begin{array}{l}S = \alpha + \beta = \frac{{11}}{2}\\P = \alpha \cdot \beta = 6\end{array} \right\} \Rightarrow {x^2} - Sx + P = {x^2} - \frac{{11}}{2}x + 6 = 0 \Rightarrow \Delta = {\left( { - \frac{{11}}{2}} \right)^2} - 4\left( 1 \right)\left( 6 \right) = 6/25\\x = \frac{{ - b \pm \sqrt \Delta }}{{2a}} = \frac{{5/5 \pm \sqrt {6/25} }}{2} \Rightarrow \left\{ \begin{array}{l}{x_1} = \frac{{5/5 + 2/5}}{2} = 4 \Rightarrow \alpha = 4\\{x_2} = \frac{{5/5 - 2/5}}{2} = 1/5 \Rightarrow \beta = 1/5\end{array} \right.\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}\alpha = \frac{{3 + \sqrt 5 }}{2}\\\beta = \frac{{3 - \sqrt 5 }}{2}\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}S = \alpha + \beta = \frac{{3 + \sqrt 5 }}{2} + \frac{{3 - \sqrt 5 }}{2} = 3\\P = \alpha \cdot \beta = \frac{{3 + \sqrt 5 }}{2} \times \frac{{3 - \sqrt 5 }}{2} = \frac{{{3^2} - {{\left( {\sqrt 5 } \right)}^2}}}{4} = 1\end{array} \right.\\ \Rightarrow {x^2} - Sx + P = 0 \Rightarrow {x^2} - 3x + 1 = 0\end{array}\)