1. questions

| ax² + bx + c | ≤ | Ax² + Bx + C | => | b²-4ac |≤ | B²-4AC |

This is a challenge question my teacher put Im only in math 102 2nd alg class, I'm a little loss I know a its polynomial theorem and the other is the discriminant. He said to treat it like a parabola and that we can remove the absolute value by just make the parabola go upwards or downwards.

First off I don't know what the difference is between the lowercase a and Uppercase A

2. Re: questions

Originally Posted by slugmar
| ax² + bx + c | ≤ | Ax² + Bx + C | => | b²-4ac |≤ | B²-4AC |

This is a challenge question my teacher put Im only in math 102 2nd alg class, I'm a little loss I know a its polynomial theorem and the other is the discriminant. He said to treat it like a parabola and that we can remove the absolute value by just make the parabola go upwards or downwards.

First off I don't know what the difference is between the lowercase a and Uppercase A
For starters, lowercase a and Uppercase A are being used to represent different parameters, since the two quadratics are different.

3. Re: questions

so can I set the them to something like this
| x² + 2x + 3 | ≤ | 2x² + 3x + 4 |

4. Re: questions

Originally Posted by slugmar
so can I set the them to something like this
| x² + 2x + 3 | ≤ | 2x² + 3x + 4 |
No, because your quadratics are arbitrary, they just happen to be different.

5. Re: questions

so If i think of them as a parabola could I set them to y= 0,c x= a,0 and b,0 and then find the vertex, Or Im I even thinking about this the right way ?

6. Re: questions

ax² + bx + c is going to have two solutions and so will Ax² + Bx + C lets say I call the solution for the first one F and G and the second H and I

ax² + bx + c
(x+F)(x+G)

Then foil out (x+F)(x+G)

x²+FX+GX+FG = ax² + bx + c

So far that is what I have done in the problem.

7. Re: questions

This is how you interpret it, if you have two quadratics, say f(x) and g(x), so that |f(x)| is less than or equal to |g(x)| for all values of x, then the absolute value of the discriminant of f(x) will also be less than or equal to the absolute value of the discriminant of g(x). That's why you can't set the a,A,b,B,c,C to any values. They have to be values that satisfy the condition that |f(x)| less than or equal to |g(x)|.

For your latest try, just be careful, because those solutions may be complex-valued (i.e. no real solutions).

Also, it might help to picture what the parabolas would look like if they satisfied the condition (i.e. one is "eating" the other)

8. Re: questions

Thank you for the info bingk and Prove It, This is my first time ever working on a proof I find it interesting but I really have no wear to go with it I'm still working on it I have till our next exam before it has to be turned in.