Question

10．已知a、b、c三數(shù)滿足方程組$\left\{\begin{array}{l}{a+b=8}\\{ab-{c}^{2}+8\sqrt{2}c=48}\end{array}\right.$，則方程bx²+cx-a=0的根是x₁=$\frac{-\sqrt{2}+\sqrt{6}}{2}$，x₂=$\frac{-\sqrt{2}-\sqrt{6}}{2}$．

Answer 1

分析 根據(jù)$\left\{\begin{array}{l}{a+b=8}\\{ab={c}^{2}-8\sqrt{2}c+48}\end{array}\right.$知a、b是方程x²-8x+c²-8$\sqrt{2}$c+48=0，即（x-4）²+（c-4$\sqrt{2}$）²=0的兩根，從而得出$\left\{\begin{array}{l}{a=b=4}\\{c=4\sqrt{2}}\end{array}\right.$，代入到方程bx²+cx-a=0求解即可．

解答 解：由題意可知$\left\{\begin{array}{l}{a+b=8}\\{ab={c}^{2}-8\sqrt{2}c+48}\end{array}\right.$，
∴可令a、b是方程x²-8x+c²-8$\sqrt{2}$c+48=0的兩根，
∴（x-4）²+（c-4$\sqrt{2}$）²=0，
∴$\left\{\begin{array}{l}{x-4=0}\\{c-4\sqrt{2}=0}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=4}\\{c=4\sqrt{2}}\end{array}\right.$，即$\left\{\begin{array}{l}{a=b=4}\\{c=4\sqrt{2}}\end{array}\right.$，
∴bx²+cx-a=0可化為4x²+4$\sqrt{2}$x-4=0，即x²+$\sqrt{2}$x-1=0，
解得：x₁=$\frac{-\sqrt{2}+\sqrt{6}}{2}$，x₂=$\frac{-\sqrt{2}-\sqrt{6}}{2}$．
故答案為：x₁=$\frac{-\sqrt{2}+\sqrt{6}}{2}$，x₂=$\frac{-\sqrt{2}-\sqrt{6}}{2}$．

點(diǎn)評(píng) 本題考查根與系數(shù)的關(guān)系，關(guān)鍵是先構(gòu)造方程，然后根據(jù)非負(fù)數(shù)性質(zhì)求出a、b、c的之后再進(jìn)行求解．