Question

11．解方程組：$\left\{\begin{array}{l}{b=2k+2}\\{^{2}=2|k|}\end{array}\right.$．

Answer 1

分析 首先把b=2k+2代入b²=2|k|，去掉絕對(duì)值化為方程組$\left\{\begin{array}{l}{4{k}^{2}+8k+4=2k}\\{4{k}^{2}+8k+4=-2k}\end{array}\right.$，求解方程組即可得答案．

解答 解：把b=2k+2代入b²=2|k|，
即（2k+2）²=2|k|，
整理得4k²+8k+4=2|k|，
去絕對(duì)值化為方程組$\left\{\begin{array}{l}{4{k}^{2}+8k+4=2k}\\{4{k}^{2}+8k+4=-2k}\end{array}\right.$，
4k²+8k+4=2k不合題意舍去．
解一元二次方程4k²+8k+4=-2k，
得k=-$\frac{1}{2}$或k=-2．
∴方程組的解為：$\left\{\begin{array}{l}{b=1}\\{k=-\frac{1}{2}}\end{array}\right.$或$\left\{\begin{array}{l}{b=-2}\\{k=-2}\end{array}\right.$．

點(diǎn)評(píng) 本題考查了如何解一元二次方程，本題的關(guān)鍵是掌握怎樣去掉絕對(duì)值，是基礎(chǔ)題．