Finding Conditions for Quadratic Equation Roots to be Outside the Unit Circle

[Lutzee]

Homework Statement

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)

Homework Equations

bx^2 + x - 1 = 0

The Attempt at a Solution

Solve quadratic eqn with the quadratic formulae

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

This gradually leads to a contradiction.

Many thanks

 

[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.

 

[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

 

[phoenixthoth]

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

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."

 

[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?

 

[phoenixthoth]

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

 

[Hurkyl]

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

 

[Lutzee]

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

 

[Hurkyl]

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

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

 

[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!)