Betweenness of Three Roots of Three Quadratic Equations :?

Take a look at the diagram provided in the link below, which visually explains why. When I get the chance, I'll come back and explain here.

Edit: Hemant, I looked at your explanation again, and actualy it's good after all. I will put down here what you offered:

Let X, Y, Z be roots of the following equations respectively:

ax² + bx + c = 0

(1/2)ax² + bx + c = 0

-ax² + bx + c = 0

We can subtract (1/2)ax² from the first and add (3/2)ax² to the last, so that those equations become:

(1/2)ax² + bx + c = -(1/2)ax² < 0

(1/2)ax² + bx + c = 0

(1/2)ax² + bx + c = (3/2)ax² > 0

because we know that a is positive. Hence we know that:

(1/2)aX² + bX + c = -(1/2)aX² < 0

(1/2)aY² + bY + c = 0

(1/2)aZ² + bZ + c = (3/2)aZ² > 0

which means that f(X) < f(Y) < f(Z). For the special case of a parabola function, that means that X < Y < Z (or X > Y > Z). Normally, one cannot draw this conclusion for an arbitrary function, but it's proper here.

I suggest that Hemant give himself the BA, because he is the one who actually provided this proof, and it's nice.

i'm going to assume that a isn't 0 so as that the equation extremely is quadratic. enable me call the roots ? and 3?. If ? = 3? = 0, then b = c = 0 and the top holds. in any different case, the two b and c could desire to be nonzero. subsequently, be conscious that a(x - ?)(x - 3?) = ax² + bx + c = 0. a(x² - 4?x + 3?²) = ax² + bx + c = 0 ==> -4a? = b and 3a?² = c. this promises b² = 16a²?², and a = c/(3?²). Substituting the latter into the former for between the climate of a supplies b² = 16a?²c/(3?²)?² = 16ac/3. finally multiply by using 3 to acquire the needed end.