samanthawooten36
(w4)^2=2w^27w4



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frx
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w as in complex a+bi

frx
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?

samanthawooten36
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It just says solve for w. Is that what you mean?

frx
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okey then I guess it's not complex

frx
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Well then you start by distrubuting the left term, do you know how to do that?

samanthawooten36
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yes? w^24^2?

frx
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Not exactly don't forget that:\[(a+b)^{2}=a^{2}+2ab+b ^{2}\]

frx
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\[w ^{2}8w+16 =2w ^{2}7w4\]

frx
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\[w ^{2}w+20=0\]

frx
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\[w ^{2}+w20=0\]

frx
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\[w= \frac{ 1 }{ 2 } \pm \sqrt{\frac{ 1^{2} }{ 2^{2}}+20}\]

frx
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\[w= \frac{ 1 }{ 2 } \pm \sqrt{\frac{ 81 }{4 }} = \frac{ 1 }{ 2 } \pm \frac{ 9 }{ 2 }\]

frx
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The two solutions which satisfies the equation are
\[w _{1}=5; w _{2}=4\]