# Roots lying between the roots of a given equation

1. Nov 1, 2012

### utkarshakash

1. The problem statement, all variables and given/known data
For what real values of 'a' do the roots of the equation $x^2-2x-(a^2-1)=0$ lie between the roots of the equation $x^2-2(a+1)x+a(a-1)=0$

2. Relevant equations

3. The attempt at a solution
The required conditions are
$\large D_1\geq0$
$\large D_2\geq0$
$\large R_1<\dfrac{-2}{2}<R_2$
where
$D_n=$Discriminant of nth equation and $R_m=$ mth roots of the latter equation

Solving simultaneously the above inequalities and taking intersection
$a\in \left(\frac{-1}{3},0\right)$
$a\in \left(\frac{-1}{4},1\right)$

2. Nov 1, 2012

### haruspex

Where does the -2/2 come from? Shouldn't there be an 'a' in there?

3. Nov 1, 2012

### utkarshakash

Sorry I made a mistake. It must be 2/2 (-b/2a)

4. Nov 2, 2012

### haruspex

But that's only the average of the roots. You need the condition that each root individually is in that range.

5. Nov 3, 2012

### utkarshakash

You are not understanding. Its not the average of the roots. Its the x-coordinate at which the minimum value of function occurs.

6. Nov 3, 2012

### haruspex

OK, but why is that interesting?

7. Nov 4, 2012

### ehild

Express the roots in terms of 'a' first. The non-negativity of the discriminants is not enough.

If you are right, both roots of the first equation lie between or equal to one of the roots of the second equation in case a=-1/3. Is it true?

ehild

8. Nov 4, 2012

### utkarshakash

I started by finding roots then substituting them for x in the latter equation and setting them <0. The answer which I get is correct. So, no worries!

9. Nov 4, 2012

### ehild

Did you get $a\in \left(\frac{-1}{4},1\right)$? That is the correct answer, I have checked. Why should the roots be negative???? Try with a=0.5. Do the roots of the first equation lie between the roots of the second one?

ehild

10. Nov 4, 2012