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solve for x. sqrt--(4x+1) +1=40
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x=380
\[is \it \sqrt{4x+1} +1 = 40....?\]
subtract 1 from the both sides &: \[4x+1=39^{2}\] 4x=39^2 -1 solve it for x, please... OK?
x=1520/4 x=380
\[\sqrt{4x+1}+1=40\] \[\sqrt{4x+1}=40-1=39\] Squaring both sides \[4x+1=1521\] 4x=1520 x=380
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if u put 380 in for x. the whole problem adds up to 40
thanks guys (:
that's right! you find the value of x that satisfies your original equation :)
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