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Math Help - Finding zero

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by cam_ddrt View Post
    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).


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  3. #3
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    Quote Originally Posted by cam_ddrt View Post
    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<br />
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