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extranous math 

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 Sep 30, 2018
 #1
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\(x+2 = \sqrt{3x+10} \\ (x+2)^2 = 3x + 10 \\ x^2 + 4x+4 = 3x + 10 \\ x^2 + x -6 = 0 \\ (x+3)(x-2) = 0 \\ x = -3, 2 \\ \text{however }\dots \text{when we plug these back in to check the solutions} \)

 

\(-3+2 \overset{?}{=} \sqrt{3(-3) + 10} \\ -1 \overset{?}{=} \sqrt{\text{of anything}} \\ \text{over the real numbers... no} \\ \\ 2+2 \overset{?}{=} \sqrt{3(2) + 10} \\ 4 \overset{?}{=} \sqrt{16} \\ \text{yes}\)

 

\(\text{So we see that x=-3 is an extraneous solution}\\ \text{arising because we squared things in order to solve the original equation} \\ \\ \text{the range of the square root function is only non-negative reals} \\ \text{and thus we need to be careful and check solutions whenever we square things to solve an equation}\)

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 Sep 30, 2018

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