# How close?

• Jan 21st 2010, 11:57 PM
l flipboi l
How close?
Hello,

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

$1/(x+3)^4 > 10,000$
• Jan 22nd 2010, 12:06 AM
drumist
$(x+3)^4$ is non-negative, so we can just move some stuff around:

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

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

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

$\implies~ \frac{1}{10} > |x+3|$
• Jan 22nd 2010, 12:24 AM
l flipboi l
Quote:

Originally Posted by drumist
$(x+3)^4$ is non-negative, so we can just move some stuff around:

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

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

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

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

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?
• Jan 22nd 2010, 12:34 AM
drumist
Quote:

Originally Posted by l flipboi l
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 $\frac{1}{(x+3)^3} > 1000$
. It's possible for $(x+3)^3$ 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 $x$ such that $x>-3$. 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:

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

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

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