# Find conditions on coefficients of a quadratic so the roots are not in unit circle

1. May 26, 2007

### Lutzee

1. The problem statement, all variables and given/known data

How to find the conditions on the coefficients of a quadratic equation for the roots to be outside the unit circle eg bx^2 + x - 1 = 0 where b is a constant How do we find the condition(s) that b must satisfy such that the roots of the quadratic lie outside the unit circle (ie modulus greater than one)

2. Relevant equations

bx^2 + x - 1 = 0

3. The attempt at a solution

We get to -1+/- * sqrt(1+4a) < -2a

Many thanks

2. May 26, 2007

### phoenixthoth

Is the question about all quadratic equations or only ones of the form bx^2+x-1=0?

If the latter, then I think you made a typo or something with your inequality -1+/- * sqrt(1+4a) < -2a. It should have b in it. Saying the modulus of the roots are greater than 1 means
$$\left\vert \frac{-1\pm \sqrt{1+4b}}{2b}\right\vert >1$$
Is it a given that b is a real number?
Try to avoid multiplying both sides of an inequality by a variable for that variable might be non-positive. You can multiply both sides by |2b| because that's always positive.

Last edited: May 26, 2007
3. May 26, 2007

### Lutzee

Thanks phoenixthoth. Yes it was a typo and yes it was the latter and yes b is real number. But how do i proceed from here? It still seems to yield a contradiction. Thanks

4. May 26, 2007

### phoenixthoth

The inequality $$\left\vert \frac{-1\pm \sqrt{1+4b}}{2b}\right\vert >1$$ can be solved and the solution is, I think, an interval.

To solve it, find the intersection of the solutions to these two:
$$\left\vert \frac{-1+\sqrt{1+4b}}{2b}\right\vert >1$$
$$\left\vert \frac{-1-\sqrt{1+4b}}{2b}\right\vert >1$$.
These two can be re-written without absolute values:
$$\frac{-1+\sqrt{1+4b}}{2b}>1$$ or $$\frac{-1+\sqrt{1+4b}}{2b}<-1$$

$$\frac{-1-\sqrt{1+4b}}{2b}>1$$ or $$\frac{-1-\sqrt{1+4b}}{2b}<-1$$.

To solve these, check out http://www.sosmath.com/algebra/inequalities/ineq06/ineq06.html especially where it says "Step 1" in bold a bit down from the top. Solve each "or" inequality separately and then your final answer is the intersection of the solution sets for the two.

If you tell me what you get for the final answer, I can say "yes that's what I got" or "no that's not what I got."

Last edited: May 26, 2007
5. May 26, 2007

### Lutzee

Thanks phoenixthoth.

Labelling the 4 cases you have identified as set out by you
a) b)

c) d)

i get the following solutions to each:

a) b < 0
b) b > 2, b >0 which implies b > 2
OR: b < 2, b < 0 which implies b < 0
c) b > 0
d) 0 < b < 2

which doesn't give me any sections of the real line common to all 4 cases. What am I doing wrong?

6. May 26, 2007

### phoenixthoth

Hmm...maybe the absolute values can't be removed that way...I'll think about it.

7. May 26, 2007

### Hurkyl

Staff Emeritus
Well, part of the problem is that you're assuming the roots are real.

8. May 27, 2007

### Lutzee

Hi Hurkyl. I am not assuming the roots are real. Thanks

9. May 27, 2007

### Hurkyl

Staff Emeritus
If you write something like
$$\frac{-1+\sqrt{1+4b}}{2b}>1$$
then you are assuming the roots are real

10. May 28, 2007

### pkleinod

bx2 + x - 1 = 0; hence, the roots are
x = (-1 +/- sqrt(1+4b))/(2b).

The surd is zero for b=-1/4, so the roots will be complex for b < -1/4 and real
for b > -1/4. For b = -1/4, x = 2, so this value of b is in the allowed range.
Consider the real and complex cases separately.

Complex roots:
-------------
Write b = -1/4 - y2, (where y > 0). and show that

|x|2 = 4/(1+4y2).

This is > 1 for 0 =< y < sqrt(3)/2, so b must be > -1.

Real roots:
----------
Write b = -1/4 + y2 and show that the roots are

2/(1-2y) and 2/(1+2y). The second of these becomes 1 for y=1/2,
i.e. b = 0. The first of these is > 1 for 0 =< y <1/2. Hence,
b must be < 0. The interval of b for which neither of the roots lies
within the unit circle is -1 < b < 0. (if my arithmetic is OK!)