{1} has an element {2}
{1, 2} has two elements.
Then let's make b equal to the above four groups.
Equal to an empty set,
express
The equation you gave has no solution. The condition that the equation has no solution is
b^2-4ac<; 0
that is
4-4(a- 1)& lt; 0
therefore
a & gt2
When it is equal to {1}, that is, x= 1, it is substituted.
a=2,
Then a=2, bring it back to verify whether X only gets 1, that is,
Is to see if the set b is equal to {1}
When it is equal to {2}, that is, x=2, substitute.
a= 1。
Then a= 1, and bring it back to verify the discovery.
x=2
or
x=0
So at this time,
Set b = {0,2}
So a= 1
unsatisfied
When it is equal to {1, 2}, it means that 1, 2 is two roots of the equation. According to Vieta theorem
We know that no matter what the value of a is,
No equation can have roots at the same time.
1,2
finally
One that is greater than or equal to
2
Supplement: Vieta Theorem: Equation A.
x^2+bx+c=0
The two roots of ""
It's x 1
x2
therefore
x 1
+x2
=
-b/a
x 1
*x2
=c/a
Give favorable comments
Hmm. How interesting