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

Algebra.Com
Question 1196478: How do you know that the equation has no real roots?

Found 3 solutions by ikleyn, greenestamps, math_tutor2020:
Answer by ikleyn(52799)   (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)   (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)   (Show Source): You can put this solution on YOUR website!

The smallest can get is 0, assuming x is a real number. Squaring a negative gets us a positive
Eg:

The same goes for and
because and
We can rewrite those terms to be in the form of

This means the smallest can get is

By extension, the smallest can get is . This is the lower bound of the range. There is no upper bound.

The range of is the the inequality which gives the interval notation [6, ∞)
Or put simply: the range is

As you can see, the output is not possible.
There are no real number solutions, or roots, for
A root is an x input that produces the output
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.
RELATED QUESTIONS

The polynomial equation x(x2+4)(x2-x-6)=0 has how many real... (answered by fractalier)
Use the discriminant to determine how many real-number roots the equation has. Do not... (answered by Earlsdon)
Determine whether the equation has two distinct real number solutions, one real numbers... (answered by checkley77)
Find k such that the equation 6x^2 + 6x + k =0 has Two equal real roots, two distinct... (answered by MathLover1,josmiceli,MathTherapy)
Find k such that the equation 2x^2 +kx + 8 = 0 has two equal real roots, two distinct... (answered by MathLover1,MathTherapy)
Find the exact number of real roots of the equation x^6-x^3+2x^2-3x-1=0... (answered by lwsshak3)
Show the equation x+2/x-3 = x+4/2x+3 has no real... (answered by Boreal)
I couldn't work letter C and #2, please assist. Please check the rest of the work and... (answered by yuendatlo,solver91311)
Find the simplest polynomial equation with the real coefficients that has the given roots (answered by rothauserc)