It is known that the quadratic equation (ab-2b) x & # 178; + 2 (B-A) x + 2A AB = 0 has two equal real roots, then the real roots are A-1 B1 C0 D-2

It is known that the quadratic equation (ab-2b) x & # 178; + 2 (B-A) x + 2A AB = 0 has two equal real roots, then the real roots are A-1 B1 C0 D-2

It can be seen from the meaning of the title: 4 (B-A) &# - 4 (ab-2b) (2a-ab) = 0
（A＋B）²－2AB（A＋B）＋A²B²＝0
（A＋B－AB）²＝0
∴ A＋B－AB＝0
A＋B＝AB
x1+x2=-2(B-A)/(AB-2B)=-2(B-A)/(A+B-2B)=2
x1x2=(2A-AB)/(AB-2B)=(2A-A-B)/(A+B-2B)=1
So the solution is X1 = x2 = 1
Select b1

Finding the fourth order determinant a 100 - 1 B 100 - 1 C 100 - 1

R1 + AR20 1 + AB a 0 - 1 B 100 - 1 C 100 - 1 D expand 1 + AB a 0 - 1 C 10 - 1 DC3 + DC21 + AB a ad - 1 C1 + cd0 - 10 expand 1 + AB ad - 1 1 + CD determinant = (1 + AB) (1 + CD) + ad = 1 + AB + CD + AD + ABCD