{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