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kandsbbf
 one year ago
Best ResponseYou've already chosen the best response.00, –2.48, 0.35 –2.48 –2.48, 0.35 0, 2.48, –0.35

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1lol once is enough ^^ sorry I'm also helping someone else so it's hard to get away XD

kandsbbf
 one year ago
Best ResponseYou've already chosen the best response.0i didn't mean too lol my cp is slow.. thought u didn't get it

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large 8x^317x^2+7x=0\]See how EVERY term has an X in it? Let's start by factoring an X out of each term.\[\large x(8x^217x+7)=0\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1I 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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Anyway doing it this way, if we apply the Zero Factor Property, we'll set each factor equal to 0 giving us,\[\large x=0 \qquad \text{and}\qquad 8x^217x+7=0\]

kandsbbf
 one year ago
Best ResponseYou've already chosen the best response.0maybe but i don't have one and their are actual graphs as the answersbut i just gave u the points.....

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1From here, we can immediately cancel out two of the options, because it has to be one of the choices with x=0.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1To find the other two roots, we can throw the quadratic part into the Quadratic Formula.\[\large ax^2+bx+c=0 \qquad \rightarrow \qquad x=\frac{b \pm \sqrt{b^24ac}}{2a}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large 8x^217x+7=0 \qquad \rightarrow \qquad x=\frac{17 \pm \sqrt{17^24(8)(7)}}{2(8)}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hopefully I plugged that in correctly. You can punch that into your calculator and hopefully get something that's listed :O
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