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第9頁(yè)

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解：在方程?$x^2-4x + 1 = 0$?中

?$a = 1，$??$b=-4，$??$c = 1$?

?$ $?兩根之和?$x_{1}+x_{2}=-\frac {-4}{1}=4，$?

兩根之積?$x_{1}x_{2}=\frac {1}{1}=1$?

解：在方程?$3x^2-\sqrt {2}x-1 = 0$?中

?$a = 3，$??$b=-\sqrt {2}，$??$c=-1$?

兩根之和?$x_{1}+x_{2}=-\frac {-\sqrt {2}}{3}=\frac {\sqrt {2}}{3}$?

兩根之積?$x_{1}x_{2}=\frac {-1}{3}=-\frac {1}{3}$?

解：?$2x^2=-3x$?

?$a = 2，$??$b = 3，$??$c = 0$?

?$ $?兩根之和?$x_{1}+x_{2}=-\frac {3}{2}，$?

兩根之積?$x_{1}x_{2}=\frac {0}{2}=0$?

解：?$4x^2=3$?

?$a = 4，$??$b = 0，$??$c=-3$?

?$ $?兩根之和?$x_{1}+x_{2}=-\frac {0}{4}=0，$?

兩根之積?$x_{1}x_{2}=\frac {-3}{4}=-\frac {3}{4}$?

解：設(shè)另一個(gè)根是?$m$?

兩根之積?$-2m=-2，$?∴?$m= 1$?

兩根之和?$-2+1=-k，$?∴?$k=1$?

綜上所述，方程的另一根是?$1 ，$??$ k$?的值是?$1$?

解：設(shè)方程的另一個(gè)根為?$t$?

∴?$t+2-\sqrt 3=4，$??$(2-\sqrt 3)t= c$?

∴?$t =2+\sqrt 3$?

∴?$c=(2-\sqrt 3)(2 +\sqrt 3)= 1$?

∴方程的另一個(gè)根為?$ 2+\sqrt 3，$??$c $?的值為?$1$?

解：?$(1)$?由題意可得?$ x_{1}+ x_{2}=- \frac {a}= \frac 52，$??$x_{1}× x_{2}= \frac {c}{a}=1$?

原式?$= x_{1}x_{2}+x_{1}+x_{2}+1=1+ \frac 52+1= \frac 92$?

?$(2)$?原式?$= \frac {x_{1}^2+x_{2}^2}{x_{1}x_{2}}= \frac {(x_{1}+x_{2})^2-2x_{1}x_{2}}{x_{1}x_{2}}= \frac {25}4-2= \frac {17}4$?

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