# Quadratic Equation Solutions

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• Sep 18th 2008, 10:12 AM
Logic
Quadratic Equation Solutions
Greetings,

$f(x)=x^2-(a^2+2a)x+a^3+a^2$ is a given function.
We are looking for all values of a such that at least one of the solutions of the function is within the interval [-1;1].

So, I've come up with this:
to find all relevent a values for exactly when one x is within [-1;1], we need this:
-1 <= x <= 0, which happens when f(-1)f(1) <= 0. (1)

To find all relevent values of a for when we need both solutions to be within [-1;1], we need:
D>=0 (2)
af(1)>=0 (3)
-1<= -b/2a <= 1 (4)

and we put it all from 1 to 4 in a system.

For the solution of (1), I have:
a is from (-1-sqrt5) / 2 to -1 U from (-1+sqrt5) / 2 to 1
(Sorry, I couldn't get Latex to work.. )

For (2), I believe all values of a do the job..

For (3):
a (-infinity;(-1-sqrt5)/2] U [1;+infinity]...

and finally for (4):
[-1-sqrt3;-1+sqrt3]

So, when we combine all of this, eventually, I am left with:
a [-1; (-1-sqrt5)/2]... which is not correct.

The true thing is supposed to be:
[(-1-sqrt5)/2; 1]

Thank you..

and sorry again for the lame writing...
• Sep 19th 2008, 09:40 AM
CaptainBlack
Quote:

Originally Posted by Logic
Greetings,

$f(x)=x^2-(a^2+2a)x+a^3+a^2$ is a given function.
We are looking for all values of a such that at least one of the solutions of the function is within the interval [-1;1].
.

Apply the quadratic formula to:

$x^2-(a^2+2a)x+a^3+a^2=0$

This gives the roots as $a$ and $a(a+1)$, so the first root is in $[-1,1]$ when a is in this interval. So all we now need to do is determine those values of $a$ for which $a(a+1)$is in $[-1,1]$.

RonL

rONl