How do you solve #x- 4\geq \frac { 8} { x - 2}#?

Question

Eva Abramson · Accepted Answer

#[0,2)uu(2,oo)# or #0<=x<2# and #x>=6# (Both are the same, just in different notation)

Multiply both sides by #x-2# to isolate the nasty fraction on the RHS. #(x-4)(x-2)>=8# Expand #x^2-6x+8>=8# Subtract #8# from both sides #x^2-6x>=0# Solve the quadratic. You can use the Quadratic Formula, but I will just solve by factorising since it is the quickest. #x(x-6)>=0# #x>=0, x>=6# Now note that in the original equation, #x=2# is undefined because the denominator equals #0#. Therefore we have a new interval - #[0,2)uu(6,oo)# which is also the same as #0<=x<2# and #x>=6# And that is our answer.

Kennedy Adams · Answer

undefined To solve the inequality $x - 4 \geq \frac{8}{x - 2}$, we can first multiply both sides of the inequality by $x - 2$ to clear the fraction:

$ (x - 2)(x - 4) \geq 8 $

Expanding the left side of the inequality:

$ x^2 - 6x + 8 \geq 8 $

Subtracting 8 from both sides:

$ x^2 - 6x \geq 0 $

Now, we can factor the left side:

$ x(x - 6) \geq 0 $

The critical points are $x = 0$ and $x = 6$. These values divide the number line into three intervals: $(- \infty, 0]$, $(0, 6]$, and $(6, \infty)$.

We can test a value from each interval to determine the sign of the expression $x(x - 6)$ within that interval:

For $x = -1$, $x(x - 6) = (-1)(-1 - 6) = (-1)(-7) = 7$, which is positive.
For $x = 3$, $x(x - 6) = (3)(3 - 6) = (3)(-3) = -9$, which is negative.
For $x = 7$, $x(x - 6) = (7)(7 - 6) = (7)(1) = 7$, which is positive.
Therefore, the solution to the inequality $x - 4 \geq \frac{8}{x - 2}$ is $x \leq 0$ or $x \geq 6$.