Help With Polynomial Inequality to The Sixth Power: x^6+9x^5+13x^4>-7x^4

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I’m solving the inequality for x. Here it is: x^6+9x^5+13x^4>-7x^4.

I factor the equation into: x^4(x+5)(x+4)>0 This means that x=-5, x=-4, and x=0 at zero. Now I must put these values on a number line.

Ok, now that these values are on the number line, I pick the test points x=-6, x=-4.5,x=-3, and x=1. The first, third, and last test points satisfy the inequality. This is where the trouble begins.

I write my answers in interval notation as follows: (-infinity,-5) and (-4,infinity). The reason why I got. (-4, infinity) and not (-4,0) and (0,infinity) is because BOTH sections satisfy the inequality, so I combined them into the larger solution (-4,infinity) because both sections satisfy the inequality. I’ve been doing this type of thing and getting the right answer for as long as i’ve been doing these types of problems, but now I’m wrong.

What is is my mistake? Is Symbolab’s solution just wrong? What can I do in the future to fix this mistake?

Thanks for all the help.

 

[pka]

I’m solving the inequality for x. Here it is: x^6+9x^5+13x^4>-7x^4.
I factor the equation into: x^4(x+5)(x+4)>0 This means that x=-5, x=-4, and x=0 at zero. Now I must put these values on a number line.

I get \(\displaystyle (-\infty,-5)\cup (-4,0)\cup (0,\infty)\), \(\displaystyle x\) cannot equal zero because \(\displaystyle 0\not > 0\)

 

[Dr.Peterson]

I write my answers in interval notation as follows: (-infinity,-5) and (-4,infinity). The reason why I got. (-4, infinity) and not (-4,0) and (0,infinity) is because BOTH sections satisfy the inequality, so I combined them into the larger solution (-4,infinity) because both sections satisfy the inequality. I’ve been doing this type of thing and getting the right answer for as long as i’ve been doing these types of problems, but now I’m wrong.

What is is my mistake? Is Symbolab’s solution just wrong? What can I do in the future to fix this mistake?

When I put the zeros on the number line, I immediately take note of whether they themselves are solutions, before looking at the intervals. In this case, since it has the form ___ > 0, zeros are not solutions, so I put open dots there. Then, after filling in the intervals where the polynomial is positive, I see that 0 is not included, and can translate the graph into the correct interval notation.

To put it another way, if you actually wrote (-infinity,-5) U (-4,0) U (0,infinity) first, you should have noted that (-4,0) U (0,infinity) is not (-4,infinity); to get that, it would have to be (-4,0) U [0,infinity).

 

[JeffM]

Find the domain such that

\(\displaystyle x^6 + 9x^5 + 13x^4 > -\ 7x^4 \iff f(x) = x^6 + 9x^5 + 20x^4 > 0.\)

Solve the related equality.

\(\displaystyle 0 = f(x) = x^6 + 9x^5 + 20x^4 = x^4(x^2 + 9x + 20) = x^4(x + 5)(x + 4) \implies \\
f(x) = 0 \text { if } x = 0,\ x = -\ 4, \text { or } x = -\ 5.\)

Now because f(x) is a continuous function, we can determine everything else by evaluating whether the specified relationship obtains at the three points of equality and four properly chosen additional test points.

\(\displaystyle f(-\ 10) = (-\ 10)^4 \{(-\ 10)^2 + 9(-\ 10) + 20) = 10^4(100 - 90 + 20) > 0 \implies \\
f(x) > 0 \text { if } x \in (-\ \infty,\ -\ 5) \ \because \ f(-\ 5) = 0.\)

\(\displaystyle f(-\ 4.5) = (-\ 4.5)^4 \{(-\ 4.5)^2 + 9(-\ 4.5) + 20) = 4.5^4(20.25 - 40.5 + 20) < 0 \implies \\
f(x) \not > 0 \text { if } x \in [-\ 5,\ -\ 4] \ \because \ f(-\ 5) = 0 = f(-\ 4).\)

\(\displaystyle f(-\ 1) = (-\ 1)^4 \{(-\ 1)^2 + 9(-\ 1) + 20) = 1(1 - 9 + 20) > 0 \implies \\
f(x) > 0 \text { if }x \in (-\ 4,\ 0) \ \because \ f(-\ 4) = 0 = f(0). \)

\(\displaystyle f(1) = (1)^4 \{(1)^2 + 9(1) + 20) = 1(1 9 + 20) > 0 \implies \\
f(x) > 0 \text { if } x \in (0,\ \infty) \ \because \ f(0) = 0.\)

So there are three sub-domains where \(\displaystyle f(x) > 0\), namely

\(\displaystyle (-\ \infty,\ -\ 5),\ (-\ 4,\ 0), \text { and } (0,\ \infty).\)

And there are two sub-domains where \(\displaystyle f(x) \not > 0\), namely

\(\displaystyle [-\ 5),\ -\ 4], \text { and } 0.\)

If you think about it a bit, you will see that you cannot join the two intervals (- 4, 0) and (0, infinity) into a single interval (- 4, infinity) because 0 is not in either of the two intervals, but is in the improperly joined interval. The problem was not with your mechanics, but with your failing to consider what the interval notation means.

By the way, notice that I picked test values that were a lot easier to deal with than some of yours.