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question about 3.23

 
 
YangBober的头像
question about 3.23
YangBober - 2020年06月23日 Tuesday 15:36
 

 so I don't get this part We also have 1+2b=a201+2b=a2≥0, so 12b−12≤b. where did you get 1+2b=a^2>=0

I am confused.can I get some help


Thank you 


 
ProfessorAreteem的头像
Re: question about 3.23
ProfessorAreteem - 2020年06月23日 Tuesday 21:48
 

Using Vieta's formulas, we see that $$x_1^2 + x_2^2 = (x_1 + x_2)^2 -2x_1 x_2 = (-a)^2 -2(b) = a^2 -2b.$$The problem tells us that we also have $x_1^2 + x_2^2 = 1$, so $a^2 -2b = 1$, that is, $1 + 2b = a^2$. Regardless of the value of $a$, we have that $a^2 \geq 0$, so $1 + 2b \geq 0$.