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16x2 – 81 = 0
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try \((4x+9)(4x-9)=0\)
as \[16x^2-81\] is the difference of two squares \(a^2-b^2=(a+b)(a-b)\)
alternatively you would write \[16x^2=81\] so \[4x=9\] or \[4x=-9\]
or even \[16x^2=81\implies x^2=\frac{81}{16}\implies x=\pm\frac{9}{4}\]
lots of ways to skin this cat
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