kdarnell08
solve the square root of 6x+1 = x-1
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kdarnell08
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|dw:1358734970247:dw|
CalebBeavers
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Ok so first square both sides to eliminate the square root
kdarnell08
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what do you mean square both sides...
CalebBeavers
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like this
\[\large \sqrt{6x+1}^{2}=(x-1)^{2}\]
on the left hand side the square root and square cancel so youve got
\[6x+1=(x-1)^{2}\]
kdarnell08
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k
CalebBeavers
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Ok so now expand the right side
kdarnell08
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k
anonymous
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you got this?
CalebBeavers
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Whataya have after you expand the right side of the equation?
CalebBeavers
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\[6x+1=(x-1)^{2}\]
\[6x+1=x^{2}-2x+1\]
\[x^{2}-8x=0\]
\[x(x-8)=0\]
\[x=0 \] and \[x=8\]
check for extraneous solutions
\[\sqrt{6(0)+1}\neq 0-1\]
\[\sqrt{6(8)+1}=8-1\]
\[\sqrt{49}=7\]
so solution is x=8