evy15
use factoring to solve (x1)^29=0



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Skaematik
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dw:1353268350386:dw

viniterranova
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First at all you should develop the expression between parentesis.

evy15
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is this using factoring? I need to use all methods

evy15
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isn't this extracting the square root

evy15
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@viniterranova do u mean foil it?

viniterranova
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If you quadratic trinomial for (x1)^2 you will find. x^2+2x+1=9, so x^2+2x8=0

evy15
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and that's for the factoring method right???

viniterranova
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Pay attetion. Just plug the x´s values +3 and 3 in the (x1)^2=9. See if this is values are valid for the expression.

viniterranova
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x^2+2x8= 0 ; (x4)(x+2)=0

evy15
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well i just wanna be sure

viniterranova
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Just solve the expression (x4)(x+2)=0.

evy15
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wait, im not there yet, so you move the 9 to the other side, and foil (x1)^2 and I got x^22x+1?

viniterranova
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Just pass the 8 for the first term of the equation and subtract it. and you got x^2+2x8= 0

evy15
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is thiscorrect

evy15
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isn't it a 2x

evy15
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@Skaematik

viniterranova
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Yes, it´s 2x. Sorry for my lack of attention. So, the expression is x^22x8=0

evy15
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and are the answers I got correct?

evy15
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@Skaematik can u check my work

evy15
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@viniterranova I saw ur work, I understand now