If the quadratic equation x ^ 2 + 2 (M + 3) x + m ^ 2 + 3 = 0 has two real roots a and B, what is the minimum value of (A-1) ^ 2 + (B-1) ^ 2?

If the quadratic equation x ^ 2 + 2 (M + 3) x + m ^ 2 + 3 = 0 has two real roots a and B, what is the minimum value of (A-1) ^ 2 + (B-1) ^ 2?

There are two real number root discriminant greater than or equal to 04 (M + 3) ^ 2-4 (m ^ 2 + 3) > = 0 (M + 3) ^ 2 - (m ^ 2 + 3) > = 06m + 9-3 > = 0m > = - 1 Veda theorem a + B = - 2 (M + 3) AB = m ^ 2 + 3 (A-1) ^ 2 + (B-1) ^ 2 = a ^ 2-2a + 1 + B ^ 2-2b + 1 = (a ^ 2 + B ^ 2) - 2 (a + b) + 2 = (a + b) ^ 2-2ab-2 (a + b) + 2 = 4 (M + 3) ^ 2-2 (m ^ 2 + 3) + 4 (

If the quadratic equation x2 + 2 (M + 3) x + M2 + 3 = 0 has two real roots α and β, then the minimum value of (α - 1) 2 + (β - 1) 2 is multiple

△=4(m+3)²-4(m²+3)=24m+24>=0m>=-1α+β=-2(m+3)αβ=m²+3(α-1)²+(β-1)²=α²+β²-2α-2β+2=(α+β)²-2αβ-2(α+β)+2=4(m+3)²-2(m²+3)+4(m+3)+2=2m²+2...

It is known that the quadratic equation x ^ 2 - | Z | x + 1 = 0 with real coefficients has real roots, then the minimum value of | Z-1 + I | is

If there are real roots, the discriminant delta = | Z | ^ 2-4 > = 0
That is: | Z | > = 2
That is, Z is the outer region of the circle with radius 2 and center of the circle as the origin
|Z-1 + I | denotes the distance between Z and point (1-I)
The minimum value is R-D, and D is the distance between the center of the circle and this point
d=√2
The minimum value is 2 - √ 2