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23.(本小题满分10分)
已知关于$x的一元二次方程x^{2} + 2x + 2m = 0$有两个不相等的实数根.
(1)求$m$的取值范围.
(2)若$x_{1}$,$x_{2}是一元二次方程x^{2} + 2x + 2m = 0$的两个实数根,且$x_{1}^{2} + x_{2}^{2} = 8$,求$m$的值.
解:
(1)$\because$一元二次方程$x^{2}+2x+2m=0$有两个不相等的实数根,$\therefore b^{2}-4ac=2^{2}-4×1×2m=4-8m＞0$,解得$m＜\frac{1}{2}$.
(2)由题意,得$x_{1}+x_{2}=-2,x_{1}\cdot x_{2}=2m$,$\therefore x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}\cdot x_{2}=4-4m=8$,解得$m=-1$.当$m=-1$时,$b^{2}-4ac=4-8m=12＞0$.$\therefore m$的值为-1.
已知关于$x的一元二次方程x^{2} + 2x + 2m = 0$有两个不相等的实数根.
(1)求$m$的取值范围.
(2)若$x_{1}$,$x_{2}是一元二次方程x^{2} + 2x + 2m = 0$的两个实数根,且$x_{1}^{2} + x_{2}^{2} = 8$,求$m$的值.
解:
(1)$\because$一元二次方程$x^{2}+2x+2m=0$有两个不相等的实数根,$\therefore b^{2}-4ac=2^{2}-4×1×2m=4-8m＞0$,解得$m＜\frac{1}{2}$.
(2)由题意,得$x_{1}+x_{2}=-2,x_{1}\cdot x_{2}=2m$,$\therefore x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}\cdot x_{2}=4-4m=8$,解得$m=-1$.当$m=-1$时,$b^{2}-4ac=4-8m=12＞0$.$\therefore m$的值为-1.
答案
解:
(1)$\because$一元二次方程$x^{2}+2x+2m=0$有两个不相等的实数根,$\therefore b^{2}-4ac=2^{2}-4×1×2m=4-8m＞0$,解得$m＜\frac{1}{2}$.
(2)由题意,得$x_{1}+x_{2}=-2,x_{1}\cdot x_{2}=2m$,$\therefore x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}\cdot x_{2}=4-4m=8$,解得$m=-1$.当$m=-1$时,$b^{2}-4ac=4-8m=12＞0$.$\therefore m$的值为-1.
(1)$\because$一元二次方程$x^{2}+2x+2m=0$有两个不相等的实数根,$\therefore b^{2}-4ac=2^{2}-4×1×2m=4-8m＞0$,解得$m＜\frac{1}{2}$.
(2)由题意,得$x_{1}+x_{2}=-2,x_{1}\cdot x_{2}=2m$,$\therefore x_{1}^{2}+x_{2}^{2}=(x_{1}+x_{2})^{2}-2x_{1}\cdot x_{2}=4-4m=8$,解得$m=-1$.当$m=-1$时,$b^{2}-4ac=4-8m=12＞0$.$\therefore m$的值为-1.
