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Function Crossing X-Axis

Date: 07/11/2003 at 08:41:14
From: Stuart
Subject: The discriminant

Find the condition that must be satisfied by k in order so that the 

   2x^2 + 6x + 1 + k(x^2 + 2)

may be positive for all real values of x.

The book gives one answer, but I have two. 
Rearrange into  ax^2 + bx + c = 0 form
   x^2(2 + k) + 6x + 1 + 2k
The graph never crosses the x-axis, so there are no values of x for 
which f(x)= 0. I.e. (2+k)x^2 + 6x + 1 + 2k = 0 has no real roots, and 
therefore b^2 - 4ac < 0.

Using b^2 - 4ac < 0

   6^2 - 4(2+k)(1+2k) < 0
     -8k^2 - 20k + 28 < 0 
         (7+2k)(4-4k) < 0

from -k^2 graph 
 f(k) < 0 
for k < -3.5
and k > 1 

but the book only gives one answer, k > 1. Why?

Date: 07/11/2003 at 09:41:07
From: Doctor Peterson
Subject: Re: The discriminant

Hi, Stuart.

You found the conditions under which the graph never crosses the x 
axis. But that could mean either that it is always positive, or 
always negative. Check the sign of y for each of your answers.

- Doctor Peterson, The Math Forum 

Date: 07/11/2003 at 19:19:17
From: Stuart
Subject: The discriminant

So does this mean that because the discriminant b^2 - 4ac < 0
can apply when the graph is always positive or always negative, only 
use the positive answer? If so, why dosen't this apply to the 
following problem (from the book):

Find the values of p for which x^2 - 2px + p + 6 is positive for all 
real values of x .

As x^2 - 2px + p  + 6 is positive for all values of x the graph of f
(x) where f(x)= x^2 -2px + p + 6 is entirely above the x axis 
i.e. the graph never crosses the x-axis so there are no real values of 
x for which f(x) = 0 i.e x^2 -2px + p + 6 = 0 has no real roots.

     b^2 - 4ac < 0
4p^2 - 4p - 24 < 0
    (p+2)(p-3) < 0

therefore x^2 - 2px + p + 6 is positive for all real values of x 
provided that -2 < p < 3.

Date: 07/11/2003 at 22:26:05
From: Doctor Peterson
Subject: Re: The discriminant

Hi, Stuart.

What you've done so far does not include actually checking the answer. 
You have simply found that when p is between -2 and 3, the 
discriminant is negative, so that f(x) is never zero. You have NOT 
shown whether it is always positive. What we can do is to take one 
value of p within this interval, say 0 for simplicity, and see what 
the f(x) is:

  f(x) = x^2 - 2px + p + 6 = x^2 - 2*0*x + 0 + 6 = x^2 + 6

That is clearly always positive, so the solution works. It could have 
turned out to be always negative, and there would be no solution to 
the problem.

I suppose this doesn't really prove that f(x) is positive for all x 
whenever -2 < p < 3; to show that we have to look at the coefficient 
of x^2 for any such p. Since it is always 1, it is positive and we 
know f(x) curves upward. A quadratic that was always negative could 
not do that, so we know the solution works.

In your problem, you found that the function would never be zero if

  k < -3.5  or  k > 1

This gives you not just one but TWO intervals to check. 

This time the coefficient of x^2 depends on k. When k < -3.5, the 
coefficient 2 + k is always negative. That means f(x) curves downward, 
and must be negative for large x. Since it never crosses the axis, it 
must always be negative.

Similarly, for k > 1, 2 + k is positive and f(x) is always positive.

To check with specific numbers, you have to try a value of k in each 

  k = -4: f(x) = x^2(2 + k) + 6x + 1 + 2k = -2x^2 + 6x - 7

We know this never crosses zero, so the fact that it is negative for 
x = 0 means that it is always negative.

  k = 2: f(x) = x^2(2 + k) + 6x + 1 + 2k = 4x^2 + 6x + 5

Here, for k > 1, f(x) is always positive.

The lesson: never treat math like a machine that will automatically 
give the answer without your having to think. Always check the answer, 
and always think about your reasoning to check for holes. In this 
case, the fact that a function never crosses the x-axis is not 
equivalent to its being always positive. As we say, it is necessary, 
but not sufficient. That means that you may have found more answers 
than are correct, and checking is essential.

- Doctor Peterson, The Math Forum 

Date: 07/12/2003 at 05:11:13
From: Stuart
Subject: Thank you (The discriminant)

Thanks for all your help. Your answers have been educational.
Associated Topics:
High School Functions

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