# How close?

#### l flipboi l

Hello,

Problem: How close to -3 do we have to take x so that

$$\displaystyle 1/(x+3)^4 > 10,000$$

#### drumist

$$\displaystyle (x+3)^4$$ is non-negative, so we can just move some stuff around:

$$\displaystyle \frac{1}{(x+3)^4} > 10,000$$

$$\displaystyle \implies~ \frac{1}{10,000} > (x+3)^4$$

$$\displaystyle \implies~ \sqrt[4]{\left(\frac{1}{10,000}\right)} > |x+3|$$

$$\displaystyle \implies~ \frac{1}{10} > |x+3|$$

l flipboi l

#### l flipboi l

$$\displaystyle (x+3)^4$$ is non-negative, so we can just move some stuff around:

$$\displaystyle \frac{1}{(x+3)^4} > 10,000$$

$$\displaystyle \implies~ \frac{1}{10,000} > (x+3)^4$$

$$\displaystyle \implies~ \sqrt[4]{\left(\frac{1}{10,000}\right)} > |x+3|$$

$$\displaystyle \implies~ \frac{1}{10} > |x+3|$$
Great thanks!

I just have a couple of rookie questions.

If it was negative, you couldn't move stuff around?

and why did the (x+3) become an absolute value?

#### drumist

Great thanks!

I just have a couple of rookie questions.

If it was negative, you couldn't move stuff around?

and why did the (x+3) become an absolute value?
Say for example we considered something like $$\displaystyle \frac{1}{(x+3)^3} > 1000$$
. It's possible for $$\displaystyle (x+3)^3$$ to be a negative number, but if it was negative, the value on the left-hand side would have been negative, and a negative number can't be greater than 1000. So we just need to make a side note that we are only considering values of $$\displaystyle x$$ such that $$\displaystyle x>-3$$. From then we can continue the problem.

As for your second question: Any time you take an even-powered root of a value, you must take its absolute value. In other words:

$$\displaystyle a^2 = 25 ~\implies~ \sqrt{a^2}=\sqrt{25} ~\implies~ |a|=5 ~\implies~ a=\pm5$$

Notice that had we not done so, we would not have found the solution $$\displaystyle a=-5$$, but clearly $$\displaystyle (-5)^2=25$$.

You do not need to do this when taking odd-powered roots.

l flipboi l