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Mathematics 8 Online
Kyky232:

x^3-34x-12=0 Do you know the solution set?

GalaxyzStarz:

As in a factored form?

Kyky232:

I dont even know. Its algebra 2

GalaxyzStarz:

\[(x \pm?)(x \pm ?)\] Like that?

Kyky232:

1 attachment
Kyky232:

thats the question

GalaxyzStarz:

Is B.) correct or incorrect?

Kyky232:

I got part B incorrect. The first part was C

GalaxyzStarz:

I need the correct answer for B.) to use on C.)

Kyky232:

the answer for b is 6, i just got it wrong the first time

surjithayer:

factors of 12 are \[\pm1,\pm2,\pm3,\pm4,\pm6\] by hit and trial method put x=1,2,3,4,5,6 when we put x=6 \[6^3-34\times6-12=216-204-12=0\] so one solution is x=6 by synthetic division 6| 1 0 -34 -12 | - 6 36 12 |


1 6 2 |0 \[x^2+6x+2=0\] \[x=\frac{ -6 \pm \sqrt{6^2-4\times1\times2} }{ 2\times2}\] \[x=\frac{ -6 \pm \sqrt{36-8} }{ 2}=\frac{ -6 \pm \sqrt{28} }{ 2 }=\frac{ -6 \pm 2\sqrt{7}}{ 2}\] \[x=-3 \pm \sqrt{7}\]

QueenNiyah:

x=6

surjithayer:

\[solution~set~is~\]\[ (-3-\sqrt{7},-3+\sqrt{7},6)\]

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