If a, B and C are three sides of △ ABC, and a and B satisfy the relation | A-3 | + (B-4) + 0, C is a system of inequalities x-3 / 3 > x-4,2x + 3x-4,2x + 3

If a, B and C are three sides of △ ABC, and a and B satisfy the relation | A-3 | + (B-4) + 0, C is a system of inequalities x-3 / 3 > x-4,2x + 3x-4,2x + 3

Because: | A-3 | + (B-4) = 0, so: A-3 = 0, B-4 = 0, so: a = 3, B = 4
Because: C is a system of inequalities x-3 / 3 > x-4,2x + 3

If a, B and C are three sides of triangle ABC, and a and B satisfy the relation (B-4) ^ 100 + | a-6 | = 0, C is a system of inequalities 4x + 6

a＝6,b＝4,2.5＜x＜5.5,c＝5 a＋bc＝26

A problem of quadratic inequality of one variable in Senior High School
Ax ^ 2 + BX + C is greater than or equal to 0, and the value range of X is: - 2

Ax ^ 2 + BX + C is greater than or equal to 0, and the value range of X is: - 2

2-4X
———— >0
5+3X

Multiply by (3x + 5) & # > 0
So (2-4x) (3x + 5) > 0
Divide both sides by - 2
(2x-1)(3x+5)