Hello,
Can you please help me figure this out?
Problem: How close to -3 do we have to take x so that
$\displaystyle 1/(x+3)^4 > 10,000$
$\displaystyle (x+3)^4$ is non-negative, so we can just move some stuff around:
$\displaystyle \frac{1}{(x+3)^4} > 10,000$
$\displaystyle \implies~ \frac{1}{10,000} > (x+3)^4$
$\displaystyle \implies~ \sqrt[4]{\left(\frac{1}{10,000}\right)} > |x+3|$
$\displaystyle \implies~ \frac{1}{10} > |x+3|$
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.