Question

Prove that x^4 + 3x^2 - 2 = 0 has exactly two solutions.

Answer 1

2? should have "4"

Answer 2

no its 2

Answer 3

however, you can reduce it to a quadratic by the substitution u=x^2
$$u^2+3u-2=0$$
solve for u
then put x back

Answer 4

huh? i though you find the derivative of x^4 + 3x^2 - 2 first

Answer 5

so, technically, there are "4" solutions...
each square-root at the end will give you two values

Answer 6

@Mertsj can u help?

Answer 7

oh.. you mean that way..

Answer 8

yea

Answer 9

no .. derivative will give you extrema.. not the roots of the function

Answer 10

@tcarroll010 can u help?

Answer 11

Did you solve it?

Answer 12

Did you graph it?

Answer 13

no

Answer 14

I'm stuck after i found the derivative

Answer 15

f'(x) = 4x^3 + 6x

Answer 16

Are you talking about real solutions or all solutions because it has 2 real and 2 complex solutions.

Answer 17

http://www.wolframalpha.com/input/?i=x^4+%2B+3x^2+-+2

Answer 18

I'm talking about rolle's theorem If f (x) is a differentiable function on the interval [a, b] such that f (a) = f (b), then there exists at least one value a < c < b such that f ′(c) = 0.

Answer 19

wait no theorom 9.2 sorry that is the wrong one

Answer 20

so?

Answer 21

Quickest and easiest way: There are 2 real solutions because of Descartes' rule of signs. If you put any positive x in, you have one sign change, so one positive real solution. If you put any negative x in, the same, fone sign change, for one negative real solution. No need to solve to get the # of solutions.

Answer 22

okay how do i do that?

Answer 23

Descartes' Rule of signs only tells the POSSIBLE number of positive and negative roots, not the actual number.

Answer 24

There is nothing to "do". Positive "x" gives one sign change. Descartes' rule of signs says that you have as the number of real solutions, the number of sign changes or that number by successively subtracting 2. Here, you have an odd number, "1" sign change, and that cannot be reduced by 2 and stay positive. So, you have one positive real solution. Similar for negative "x". So, a total of 2 real solutions.

Answer 25

The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by a multiple of 2. Multiple roots of the same value are counted separately.

Answer 26

In this case, we actually do get the actual number of solutions, because we are given "1" for positive "x" and for negative "x". This is the exceptional case where Descartes' rule gives the actual number.

Answer 27

ok

Answer 28

http://www.millersville.edu/~bikenaga/courses/161/homework/ps19/ps19.pdf

Look at number 9

Answer 29

okay

Answer 30

Descartes' Rule is usually used to give the possible number of signs, but here, for this type of question, it can be used simply and easily and accurately to give the actual number. Descartes' rule is usually not used this way to give an actual number, but it works for the case of "one sign change" and "one sign change" for positive/negative.

Answer 31

Okay

Answer 32

Did you look at the solution?

Answer 33

yes i did

Answer 34

Does it seem right to you?

Answer 35

Yea

Answer 36

Good.

Answer 37

So, if all you want is to prove that there are 2 real solutions, then just look at Descartes (for this type of problem). You catch a break because of the 4th and 2nd power. It's the one legitimate "shortcut" you can derive from Descartes. It's a good technique to remember for test-taking.

Answer 38

okay

Answer 39

so i don't need to subtititue anything?

Answer 40

No, you're done by using Descartes. For this type of problem. No need to over-complicate by solving. You can go ahead and solve if you want to, but your question is not asking for that. It's good to remember: "necessary and sufficient". You don't have to do more work than is necessary.

Answer 41

okay

Answer 42

Solving is sufficient, but not necessary (it's "overkill"). Using Descartes (here) is sufficient because it does what you are asked to do without doing more than you need. Almost all the time you would try Descartes for similar questions, it would not be enough. But for this problem, it works marvelously.

Answer 43

Okay got you

Answer 44

It's almost sort of a "trick" question. What the questioner is really asking is if you can see some of the workability behind certain theorems and use them in ways not normally intended. In, short, it's a totally legitimate shortcut. This question comes up once in your lifetime.

Answer 45

okay..

Answer 46

So, all good now? @onegirl

Nice working with you again.

Answer 47

thanks

Answer 48

Good luck to you in all of your studies and thx for the recognition! @onegirl

And you're quite welcome!

Answer 49

so is this correct?

If f(x) = x4 +3x2 −2, then f(−1) = 2, f(0) = −2, and f(1) = 2. By the Intermediate Value Theorem,
there must be roots between −1 and 0 and between 0 and 1. Thus, the equation has at least two solutions.
f′(x) = 4x3 + 6x. So
0 = f′(x) = 4x3 + 6x = 2x(2x2 + 3), which gives x = 0.
(2x2 + 3 = 0 has no solutions.) Thus, there is a single horizontal tangent.
Rolle’s theorem says that there is a horizontal tangent between any two roots. Suppose there were more
than two roots. Then there are at least three roots a, b, and c — say a < b < c. By Rolle’s theorem, there
are horizontal tangents between a and b and between b and c. This contradicts the fact that there is only
one horizontal tangent.
Therefore, x4 + 3x2 − 2 = 0 has exactly two solutions.

Answer 50

Yes, that also works, but notice how much work is required that way. If you get this question on a test, if I were doing it and were pressed for time, I would look for the easiest and fastest way that still answers the question.

Answer 51

okay so can u simplify it?

Answer 52

Do you mean the way I did it?

Answer 53

yea

Answer 54

Ok. You'll be amazed at how easy my solution is. It almost looks like cheating, but it's not! Descartes says that if you are looking for real solutions, you are going to have x either negative or "0" or positive. That is rather obvious. We can see that "0" doesn't work, just by trying it, so we rule out "0". If we look for all positive solutions, we notice that by substituting in ANY positive "x", there will be exactly one sign change. Descartes tells us that the possible number of positive solutions will be the number of sign changes or some positive integer successively reduced by 2 (at a time). So, if our sign changes for positive "x" were "5", then our possible number of solutions would be either 5 or 3 or 1. Now, in our case, we have exactly one sign change, so we know absolutely that there is exactly one positive real solution. We weren't asked to find out what it is, just to come up with the number of "x". You do the same for negative "x" and see ONE sign change. So, you have a total of "2".

Answer 55

Okay that makes sense

Answer 56

It truly is a legitimate answer and methodr this particular problem. Because we came up with "1" for both positive and negative "x". If we had come up with other numbers, then we could not use Descartes for the answer.

Answer 57

okay

Answer 58

thanks again

Answer 59

Now you can "wow" your teacher!

Answer 60

lol

Answer 61

electrokid was on the right track, but he got derailed after a bit. He is right in pointing out that you can make the substitution of:

u = x^2

but that will still lead to only 2 solutions, not 4:\[u ^{2} + 3u - 2 = 0\]gives\[u = \frac{ -3 \pm \sqrt{17} }{ 2 }\]but you can't use the negative sign here because for real solutions, you can't have x^2 = any negative number, so for the "u" substitution equation, you really only have:\[u = \frac{ -3 + \sqrt{17} }{ 2 }\]because that is now positive and now for "x", you can take the positive and negative square root of that for both of your "x" values. That gives you a bit more concreteness to your answer.\[x = \pm \left( \frac{ -3 + \sqrt{17} }{ 2 } \right)\] .

Answer 62

You can use the "negative sign" in the solution for "u" to get the other 2 (complex) solutions. But of course, they are complex, not real.

Answer 63

okay thanks again! :)

Answer 64

Here's a graph of the 2 real solutions:

Answer 65

ok

Answer 66

I forgot to put the second radical in that solution:\[x = \pm \sqrt{\frac{ -3 + \sqrt{17} }{ 2 }}\]Sorry about that.

Answer 67

ok