# Find the Domain of this irrational expression

While finding the Domain of this expression $$\sqrt{(2x-8)(x-1)}$$

I got this:

$$(2x-8)(x-1)≥0$$

$$2x-8\geq0 \vee x-1\geq0$$

$$x\geq4 \vee x\geq 1$$

So the Domain is: $$x\in [1, +\infty)$$

But the real domain is

$$x\in (-\infty,1] \cup [4,+\infty)$$

So it means that $$x\leq1$$

• You only considered $x\geq4 \vee x\geq 1$ where both factors are positive; but the product is positive also if both factors are negative. Mar 2 at 8:59
• @Davide Pardon me, but still don't get it (⁠╥⁠﹏⁠╥⁠) Mar 2 at 9:05
• $(2x-8)(x-1)\ge0$ implies that $x\le1\vee x\ge4$. Mar 2 at 9:11
• Davide means that the radicand $(2x-8)(x-1) = 2x^2 - 10x + 8$ as a whole should be positive, which is different than imposing $2x-8 \geq 0$ and $x-1 \geq 0$ separately. Mar 2 at 9:12

$$(2X-8)(x-1)≥0$$

$$2x-8\geq0 \vee x-1\geq0$$

An error occurs here. In writing $$\lor$$, you are saying "or," but this is not the case for this. In general, $$ab \ge 0$$ only if both are positive, both are negative, or at least one is zero. Hence, you would instead write

$$\Big( 2x - 8 \ge 0 \text{ and } x-1 \ge 0 \Big) \text{ or } \Big( 2x - 8 \le 0 \text{ and } x-1 \le 0 \Big)$$

For instance, by writing "or," in your original expression, if $$2x-8 > 0$$ but $$x-1 < 0$$, then you get a negative under the radical -- not good!

In the first case, you can see that $$x \ge 4$$; in the second, $$x \le 1$$. Why? In the first, you derive

• $$x \ge 4$$, and
• $$x \ge 1$$

Which values $$x$$ satisfy both? Well, all $$x \ge 4$$. Likewise, for the other, we derive

• $$x \le 4$$, and
• $$x \le 1$$

The values that satisfy both are $$x \le 1$$. Hence, we conclude that the domain of the function is all $$x$$ such that $$x \ge 4 \text{ or } x \le 1$$

I suspect your original error comes from using the "property"

$$\sqrt{ab} = \sqrt a \sqrt b$$

$$\sqrt{(2x-8)(x-1)} = \sqrt{2x-8} \sqrt{x-1}$$
but this does not hold all of the time! In the first I gave with $$a,b$$, we require $$a \ge 0$$ and $$b \ge 0$$, because otherwise the square roots on the right side could have negatives in their radicals -- not good! Likewise, $$2x-8$$ and $$x-1$$ could both be negative, making the expressions on the right side above ill-defined, but their product could be positive, keeping the original function well-defined just fine.
• You meant to say that $ab \geq 0$ if both $a$ and $b$ are positive, both $a$ and $b$ are negative, or at least one of the numbers $a$ and $b$ is equal to zero. Mar 2 at 13:16
you can draw a sign table to get $$D_f$$ Domain fo the function