Peter G. answered • 09/30/16

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This is a continuous function. Plugging in 1 we get 41+C. Plugging in -1 we get the same, 41+C. Plugging in 0 we get C.

Choose C < 0 and C > -41. Then by the intermediate value theorem of Calculus, as the function goes from 41+C at x=-1 to C at 0 it crosses the x-axis, and it crosses it again as it goes from 0 to 41+C at x=1. That gives (at least) two real roots.

If you mean plural "values", i.e. you want to characterize all of the values of C that work, then certainly C > 0 doesn't work because all the other summands are perfect squares, hence are non-negative, so adding them gives something >= 0, and then adding C will take us above 0. So C <= 0 is an upper bound. We show that any value of C < 0 works. Let C be < 0. Then choose x to be the sixth (positive) root of -C. For example, if C = -729 then take x to be 3. We conclude

x

^{6}+9x^{4}+26x^{2}+C >^{ }x6 + C >= 0, and likewise taking x to be the negative square root of -C, so the intermediate value theorem argument works again, giving at least two roots.The function has one real root if C = 0.

Therefore

the function will have one or more real roots if and only if C <= 0

the function will have exactly one real root if and only if C = 0

It is also easy to show that the function has zero, one, or exactly two real roots by considering the first derivative, which is positive to the right of 0 and negative to the left of 0, hence there are no local extrema other than the minimum at 0.

Peter G.

A real root is when the graph of the function crosses the x axis. I said in the second paragraph that -2 is an answer, just not the only one. However, we don't know what you're being taught in the course, so we don't know whether to use calculus or a graphing calculator or what.

I would recommend cutting and pasting your polynomial into the graphing calculator at desmos.com. Then try using -2 for C, and you will see that it crosses the x axis . So does -1, -3, -2.5, etc.

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