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kandsbbf

  • 2 years ago

Find the real solutions of the equation by graphing. –8x3 – 17x2 + 7x = 0

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  1. kandsbbf
    • 2 years ago
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    @zepdrix

  2. kandsbbf
    • 2 years ago
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    0, –2.48, 0.35 –2.48 –2.48, 0.35 0, 2.48, –0.35

  3. kandsbbf
    • 2 years ago
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    @zepdrix

  4. zepdrix
    • 2 years ago
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    lol once is enough ^^ sorry I'm also helping someone else so it's hard to get away XD

  5. kandsbbf
    • 2 years ago
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    i didn't mean too lol my cp is slow.. thought u didn't get it

  6. zepdrix
    • 2 years ago
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    \[\large -8x^3-17x^2+7x=0\]See how EVERY term has an X in it? Let's start by factoring an X out of each term.\[\large x(-8x^2-17x+7)=0\]

  7. zepdrix
    • 2 years ago
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    I know they said to solve by GRAPHING, but I'm not exactly sure how you would go about doing that. Do they mean by using a calculator?

  8. zepdrix
    • 2 years ago
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    Graphing Calculator*?

  9. zepdrix
    • 2 years ago
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    Anyway doing it this way, if we apply the Zero Factor Property, we'll set each factor equal to 0 giving us,\[\large x=0 \qquad \text{and}\qquad -8x^2-17x+7=0\]

  10. kandsbbf
    • 2 years ago
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    maybe but i don't have one and their are actual graphs as the answersbut i just gave u the points.....

  11. kandsbbf
    • 2 years ago
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    okayy

  12. zepdrix
    • 2 years ago
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    From here, we can immediately cancel out two of the options, because it has to be one of the choices with x=0.

  13. zepdrix
    • 2 years ago
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    To find the other two roots, we can throw the quadratic part into the Quadratic Formula.\[\large ax^2+bx+c=0 \qquad \rightarrow \qquad x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

  14. zepdrix
    • 2 years ago
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    \[\large -8x^2-17x+7=0 \qquad \rightarrow \qquad x=\frac{17 \pm \sqrt{17^2-4(-8)(7)}}{2(-8)}\]

  15. zepdrix
    • 2 years ago
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    Hopefully I plugged that in correctly. You can punch that into your calculator and hopefully get something that's listed :O

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