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anonymous

  • one year ago

What are the zero(s) of the function f(x) = the quantity of 5 x squared minus 25 x, all over x?

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  1. anonymous
    • one year ago
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    @mathstudent55

  2. anonymous
    • one year ago
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    @xapproachesinfinity

  3. campbell_st
    • one year ago
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    \[\frac{5x^2 - 25x}{x}\] what common factor do you see..?

  4. anonymous
    • one year ago
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    5x

  5. campbell_st
    • one year ago
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    ok... so I'll do that \[\frac{5x(x - 5)}{x}\] what common factor do you see in the numerator and denominator..?

  6. anonymous
    • one year ago
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    x

  7. campbell_st
    • one year ago
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    ok so that leaves 5(x -5) = 0 so what value of x makes the equation true..?

  8. anonymous
    • one year ago
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    5, 0

  9. campbell_st
    • one year ago
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    only 5 and when x = 0 in the original equation you have a zero denominator, which is undefined so at x = 0 you have a point of discontinuity... so the solution is x = 5 which comes from the simplified version of the original equation. hope it makes sense.

  10. anonymous
    • one year ago
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    It does! Thank-you:)

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