SOLUTION: How do you know that the equation has no real roots? {{{ x^6 + 3x^4 + 30x^2 +6 = 0 }}}

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Question 1196478: How do you know that the equation has no real roots?
+x%5E6+%2B+3x%5E4+%2B+30x%5E2+%2B6+=+0+
Found 3 solutions by ikleyn, greenestamps, math_tutor2020:
Answer by ikleyn(52797) About Me  (Show Source):
You can put this solution on YOUR website!
.
Left side expression is the sum of four addends.


Three of these addends are monomials of even degree of variable x 
with positive coefficients - so these addends are non-negative at any real value 
of the variable x.


The fourth addend is positive constant "6".


The sum of three non-negative values and fourth positive value can not be equal to zero - 

THEREFORE, left side of the equation can not be zero at any value of x.


So, the equation has no real solutions.

At this point, the proof is complete.



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Almost certainly the explanation by tutor @ikleyn is the easiest way to show that the given polynomial has no real roots.

But note that Descartes' rule of signs can also be used.

The given polynomial has 0 sign changes, so the number of positive real roots is 0.

The polynomial with x replaced by -x also has 0 sign changes, so the number of negative real roots is also 0.

Along with the fact that x=0 is obviously not a root, that means the number of real roots is 0.

Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

The smallest x%5E2 can get is 0, assuming x is a real number. Squaring a negative gets us a positive
Eg: x%5E2+=+%28-7%29%5E2+=+%28-7%29%2A%28-7%29+=+49

The same goes for x%5E4 and x%5E6
because x%5E4+=+%28x%5E2%29%5E2 and x%5E6+=+%28x%5E3%29%5E2
We can rewrite those terms to be in the form of %28something%29%5E2

This means the smallest x%5E6%2B3x%5E4%2B30x%5E2 can get is 0%2B0%2B0=0

By extension, the smallest x%5E6%2B3x%5E4%2B30x%5E2%2B6 can get is 0%2B6+=+6. This is the lower bound of the range. There is no upper bound.

The range of y+=+x%5E6%2B3x%5E4%2B30x%5E2%2B6 is the the inequality which gives the interval notation [6, ∞)
Or put simply: the range is y+%3E=+6

As you can see, the output y+=+0 is not possible.
There are no real number solutions, or roots, for x%5E6%2B3x%5E4%2B30x%5E2%2B6+=+0
A root is an x input that produces the output y+=+0
Real numbered roots correspond to the x intercepts.

Visual confirmation with a graph
https://www.desmos.com/calculator/tcrnn8advc
The curve does not cross the x axis, so it doesn't have any x intercepts.
The curve may look like a parabola, but it's not.