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1990年全国初中数学联赛第一试有一题:方程7x~3-(K+13)x+k~(2)-k-2=0(k 是实数)有两个实根α、β,且0<α<1,1<β<2。那么 K 的取值范围是什么?解此题时,许多同学出现了下列错误解法:解:∵0<α<1,1<β<2,∴1<α+β<3,0<αβ<2.根据韦达定理α+β=(K+B)/7,αβ=(k~(2)-k-2)/7依题意有(k+13)~(2)-4·7·(k~(2)-k-2)>01<(k+13)/7<30<(k~(2)-k-2)/7<2
In the first trial of the National Junior High School Mathematical League in 1990, there was a question: Equation 7x~3-(K+13)x+k~(2)-k-2=0 (k is a real number) has two real roots alpha, beta, and 0<α<1,1<β<2. Then what is the range of K? When solving this question, many students have the following error solution: Solution: ∵0<α<1,1<β<2, ∴1<α+β<3,0<αβ< 2. According to the Vedic theorem α+β=(K+B)/7, αβ=(k~(2)-k-2)/7 depends on (k+13)~(2)-4·7 ·(k~(2)-k-2)>01<(k+13)/7<30<(k~(2)-k-2)/7<2