第9页 - 苏科版数学补充习题九年级上下册答案

解:(1) ${x}_{1}$+ ${x}_{2}$=- $\frac{b}{a}$=4

${x}_{1}$× ${x}_{2}$= $\frac{c}{a}$=1

解:(2) ${x}_{1}$+ ${x}_{2}$=- $\frac{b}{a}$= $\frac{\sqrt{2}}{3}$

${x}_{1}$× ${x}_{2}$= $\frac{c}{a}$= $-\frac{1}{3}$

解:(3) ${x}_{1}$+ ${x}_{2}$=- $\frac{b}{a}$= $-\frac{3}{2}$

${x}_{1}$× ${x}_{2}$= $\frac{c}{a}$=0

解:(4) ${x}_{1}$+ ${x}_{2}$=- $\frac{b}{a}$=0

${x}_{1}$× ${x}_{2}$= $\frac{c}{a}$= $-\frac{3}{4}$

解:设另一个根是m

两根之积-2m=-2,

所以m= 1

两根之和-2+1=-k

所以k=1

综上所述,方程的另一根是1 , k的值是1

解:设方程的另一个根为t

所以 $t+2-\sqrt{3}=4，(2-\sqrt{3})t= c$

所以 $t =2+\sqrt{3}$

所以 $c=(2-\sqrt{3})(2 +\sqrt{3})= 1$

所以方程的另一个根为 $2+\sqrt{3}$,c的值为1

解:(1)由题意可得 ${x}_{1}$+ ${x}_{2}$=- $\frac{b}{a}$= $\frac{5}{2}$

${x}_{1}$× ${x}_{2}$= $\frac{c}{a}$=1

原式= ${x}_{1}{x}_{2}+{x}_{1}+{x}_{2}+1$

=1+ $\frac{5}{2}$+1

= $\frac{9}{2}$

(2)原式= $\frac{{x}_{1}²+{x}_{2}²}{{x}_{1}{x}_{2}}$

= $\frac{({x}_{1}+{x}_{2})²-2{x}_{1}{x}_{2}}{{x}_{1}{x}_{2}}$

= $\frac{25}{4}$-2

= $\frac{17}{4}$