It is known that a > b > C, and 2A + 3B + 4C = 0. (1) proof: a + B + C > 0 It is known that a > b > C and 2A + 3B + 4C = 0 (1) Is a + B + C a positive number? Why? (2) Can the length of the line cut by the parabola y = ax ＾ 2 + BX + C on the x-axis be equal to (root 91) / 6? If you can, find out the equation of symmetry axis of parabola; if not, please explain the reason The solution of 1 2a+3b+4c=(2a+c)+3b+3c Because a > C, 2A + C

It is known that a > b > C, and 2A + 3B + 4C = 0. (1) proof: a + B + C > 0 It is known that a > b > C and 2A + 3B + 4C = 0 (1) Is a + B + C a positive number? Why? (2) Can the length of the line cut by the parabola y = ax ＾ 2 + BX + C on the x-axis be equal to (root 91) / 6? If you can, find out the equation of symmetry axis of parabola; if not, please explain the reason The solution of 1 2a+3b+4c=(2a+c)+3b+3c Because a > C, 2A + C

√ (b ^ 2-4ac) / (absolute value of a) = √ 91 / 6
(b^2-4ac)/a^2=91/36
36(b^2-4ac)=91a^2
2a+3b=-4c
36[b^2+a(2a+3b)]=91a^2
36b^2+72a^2+108ab=91a^2
19a^2-108ab-36b^2=0
a=(108b+120b)/38=228b/38=6b
x=-b/2a=-1/12
Or a = (108b-120b) / 38 = - 12b / 38 = - 6B / 19
x=-b/2a=19/12
The equation of symmetry axis of parabola is: x = - 1 / 12 or x = 19 / 12