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### Conditions for Real-valued Roots

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Date: 04/01/98 at 13:32:19
From: Heather
Subject: Calculus

Given the equation ax^3 + bx^2 + c = 0, use the first and second
derivatives to help find conditions that ensure the equation has 3
distinct real-valued roots.

So far I've figured this out....

f'(x) = 3ax^2 + 2bx
= x(3ax + 2b)
= 3ax + 2b = 0
3ax = -2b

-2b
x = ---- or 0
3a

f''(x) = 6ax + 2b = 0
6ax = -2b

-b
x = ---
3a

Also, a does not equal 0, and b does not equal 0; and f'(x) is a
parabola and f''(x) is a line.

So where do I go from here?
```

```
Date: 04/01/98 at 16:41:41
From: Doctor Rob
Subject: Re: Calculus

f(x) has three real roots if and only if

g(x) = f(x)/a,
= x^3 + (b/a)*x^2 + c/a,
= x^3 + B*x^2 + C,

has three real roots. We know that for large enough positive x,
g(x) > 0, and for small enough negative x, g(x) < 0.

You have found the two critical points x = 0, y = C, and x = -2*B/3,
y = (4*B^3 + 27*C)/27. Call these points P and Q, respectively. At
these points, g''(0) = 2*B, and g''(-2*b/3) = -2*B. Thus one is a
local maximum, the other a local minimum, by the appropriate second
derivative test. A root means that the curve y = g(x) crosses the
axis, that is, changes sign. Moving from large values of x where
g(x) > 0, to smaller values, we must cross the axis, reach a local
minimum, cross the axis again, reach a local maximum, then cross the
axis a third time, before reaching the very small values of x for
which g(x) < 0 is guaranteed. There are two cases, depending on the
sign of B.

Case 1: B > 0. Then P is the local minimum and Q is the local
maximum. Then there must be a zero between P and +infinity,
one between P and Q, and one between Q and -infinity. That
means that C < 0 and (4*B^3+27*C)/27 > 0. The last condition
is equivalent to 4*B^3 + 27*C > 0, or C > -4*B^3/27.
Thus B > 0 > C > -4*B^3/27.

Case 2: B < 0. Then Q is the local minimum and P is the local
maximum. Then there must be a zero between Q and +infinity,
one between P and Q, and one between P and -infinity. That
means that C > 0 and (4*B^3+27*C)/27 < 0. The last condition
is equivalent to 4*B^3 + 27*C < 0, or C < -4*B^3/27.
Thus B < 0 < C < -4*B^3/27.

In either case, B and C must have opposite signs, and B and
4*B^3 + 27*C must have the same sign.

In terms of a, b, and c,

b > 0 > c > -4*b^3/27*a^2, or
b < 0 < c < -4*b^3/27*a^2.

-Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 04/01/98 at 16:56:24
From: Doctor Anthony
Subject: Re: Calculus

The method is as follows.

First find conditions for the quadratic f'(x) to have real, distinct
roots.

f'(x) = 3ax^2 + 2bx = 0    so x = 0 and  x = -2b/3a

Now you require the two turning points to be on opposite sides of the
x axis, so (assuming a and b have same sign) we require

f(-2b/3a) > 0  and f(0) < 0    so c < 0

If a and b have opposite signs, we require c > 0.

a(-2b/3a)^3 + b(-2b/3a)^2 + c > 0

-8b^3/(27a^2) + 4b^3/(9a^2) + c > 0

(4/27)(b^3/a^2) + c > 0

4b^3 + 27a^2*c > 0

If a and b have the same sign, then c < 0 and we require  b > 0 and

|4b^3| > 27a^2*c

If a and b have opposite signs, then c > 0  and

4b^3 + 27a^2*c < 0

So we require b < 0 and |4b^3| > 27a^2*c

We can summarize as follows:

if a and b same signs then c < 0 and |4b^3| > 27a^2*c

if a and b opposite signs then b < 0, C > 0, and
|4b^3| > 27a^2*c

These would be sufficient conditions for three real roots. There is no
need to use the second derivative as the above are sufficient. There
is a point of inflection, given by f''(x) = 0, and this occurs at
x = -b/(3a), i.e. halfway between the maximum and minimum points.

-Doctor Anthony, The Math Forum
Check out out web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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