Question

This is a sample question from my lesson.
f(x) = √(20 - 2x)
the lesson says the answer is x ≤ 10.
In my textbook it says "the expression under a radical may not be negative."
So, how do I change the sign and solve to get the answer above?

Answer 1

You know that you can NOT take the square root of a NEGATIVE.

Correct?

Answer 2

yes, i know

Answer 3

Ok, so therefore the part inside the square root (the 20-2x), must be greater than or equal to 0.

i.e: \(20-2x\ge0\)

Answer 4

\(\color{blue}{1)}\) Add \(2x\) to both sides.
\(\color{blue}{2)}\) Divide by \(2\) on both sides.

Answer 5

Oh I see! So the addition would take away the negative sign leaving us to solve for x as normal? and adding on the other side wouldn't change ≥ to ≤?
like 20 - 2x ≥ 0
20 ≤ 2x
20/2 ≤ 2x/2
10 ≤ x?

Answer 6

incorrect

Answer 7

Follow me carefully.

\(\large\color{black}{ \displaystyle 20-2x\ge0 \\[0.5em]}\)
\(\large\color{black}{ \displaystyle 20-2x \color{blue}{+2x}\ge0\color{blue}{+2x} \\[0.5em]}\)

The \(-2x\) and \(+2x\) will cancel each other out.

\(\large\color{black}{ \displaystyle 20\ge2x \\[0.5em]}\)

Divide both sides by 2

\(\large\color{black}{ \displaystyle 10\ge x \\[0.5em]}\)

Answer 8

So, the 10 is greater than (or equal to) x. ` (Not the other way around) `

Answer 9

oh i see! i get it now! thank you so much!

Answer 10

Ok, very nice.

Answer 11

You can logically check that x is NOT greater than 10, because if you choose for example x=11, then you get:

20-2(11)
20-22
-2

and thus, since the part inside the square root can NOT be negative,, you can tell that x is NOT greater than 10.

Answer 12

And you can logically see why x=10, or other x values that are smaller than 10 make the square root 0 or greater than 0 (which is perfectly valid for us).

Answer 13

i never knew to check it like that. That's very good to know! thanks :)