Question

1.设x1,x2是方程(a-1)x^2-(2a-3)x+a=0的两根,且x1^2+x2^2=9,求a值!

Answer 1

1.x1+x2=(2a-3)/(a-1),x1*x2 = a/(a-1)

x1^2+x^2 = (x1+x2)^2-2x1x2 = 9->9(a-1)^2 = (2a-3)^2-2a = 4a^2-14a+9 = 9a^2-18a+9->5a^2-4a=0->a =0或a = 4/5

又b^2-4ac>0->4a^2-12a+9-4a^2+4a>0->-8a+9>0->a