Given that x ≠5, is x > 1/(x-5)^2?

[VJesus12]

Given that \(x \ne 5\), is \(x> \frac{1}{(x-5)^2}\)

(1) x > 0

(2) x > 10

[spoiler]OA=B[/spoiler]

Source: Magoosh

[Vincen]

Given that \(x \ne 5\), is \(x> \frac{1}{(x-5)^2}\)

(1) x > 0

(2) x > 10

[spoiler]OA=B[/spoiler]

Source: Magoosh

Hi Vjesus12.

First, let's use that \((x-5)^2 > 0\) to rewrite the given inequality as $$x>\frac{1}{\left(x-5\right)^2}\ \ \ \Rightarrow\ \ \ \ x\left(x-5\right)^2>1.$$ Now, we need to see if the inequality above is true or not.

Statement 1:

(1) x > 0

Let's try some values:

- If x=1 then \(x\left(x-5\right)^2\ =\ 1\left(-4\right)^2=16>1\), in this case the answer is YES.
- If x=0.02 then \(x\left(x-5\right)^2\ =\ 0.02\left(4.98\right)^2=0.02\left(24.8004\right)\approx0.496\), in this case the answer is NO.

So, this statement is NOT SUFFICIENT.

Statement 2:

(2) x > 10

In this case, we have that $$x-5>5\ \ \ \Rightarrow\ \ \ \left(x-5\right)^2>25\ \ \Rightarrow\ \ x\left(x-5\right)^2>x\cdot25>10\cdot25=250\ >1.$$ Hence, this statement is SUFFICIENT.

Therefore, the correct answer is the option _B_.

I hope it is clear. <i class="em em---1"></i>