chelbell409
solve 81+4x^2=0 by factoring



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abhorsen
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Remember the rule about factoring differences of squares: \(a^2b^2 = (a+b)(ab)\). Does that help you out?

chelbell409
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no can u explain it farther

abhorsen
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So are the two parts of the equation (\(81\) and \(4x^2\)) squares?

chelbell409
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ok

abhorsen
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Now set \(a=4x^2\) and \(b=81\), and plug it in to the equation I gave above.

abhorsen
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Uh... correction: \(a^2 = 4x^2\) and \(b^2 = 81\).

Kynosis
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Yep. (2x  9) (2x + 9)

quannie74
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\[81+4x^2=0\]

quannie74
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can u explain step by step because i am on these problems in my math class