1. ## Finding zero

I cannot figure out how to do this. Any help would be appreciated. Thanks!

1. Find all of x for which (x2 -x)(2x+1) - (x2 +x)(2x-1) is zero.

2. Find the domain of the function f(x) =square root of: x2 -3x. Express your answer in terms of intervals.

2. Originally Posted by cam_ddrt
2. Find the domain of the function f(x) =square root of: x2 -3x. Express your answer in terms of intervals.
You mean $f(x) = \sqrt{x^2 - 3x}$?
Only non-negative numbers can be under the square root, so solving the inequality $x^2 - 3x \geq 0$ would give us the values of x we want. Solving the corresponding equation $x^2 - 3x = 0$ would give use the critical values of 0 and 3. Now test different values of x to see if what's underneath the square root is non-negative.

For $x < 0$, $x^2 - 3x > 0$, check.
For $x = 0$, $x^2 - 3x = 0$, check.
For $0 < x < 3$, $x^2 - 3x < 0$, NOPE.
For $x = 3$, $x^2 - 3x = 0$, check.
For $x > 3$, $x^2 - 3x > 0$, check.

So our domain is $(-\infty, 0] \cup [3, \infty)$.

01

3. Originally Posted by cam_ddrt
1. Find all of x for which (x2 -x)(2x+1) - (x2 +x)(2x-1) is zero.
Multiply and simplify

$(x^2 -x)(2x+1) - (x^2 +x)(2x-1)=0$

$2x^3+x^2-2x^2-x-(2x^3-x^2+2x^2-x)=0$

$2x^3+x^2-2x^2-x-2x^3+x^2-2x^2+x=0$

$-2x^2=0$

$x=0
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