anonymous
  • anonymous
solve the square root of 6x+1 = x-1
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
Ok so first square both sides to eliminate the square root
anonymous
  • anonymous
what do you mean square both sides...

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anonymous
  • anonymous
like this \[\large \sqrt{6x+1}^{2}=(x-1)^{2}\] on the left hand side the square root and square cancel so youve got \[6x+1=(x-1)^{2}\]
anonymous
  • anonymous
k
anonymous
  • anonymous
Ok so now expand the right side
anonymous
  • anonymous
k
anonymous
  • anonymous
you got this?
anonymous
  • anonymous
Whataya have after you expand the right side of the equation?
anonymous
  • anonymous
\[6x+1=(x-1)^{2}\] \[6x+1=x^{2}-2x+1\] \[x^{2}-8x=0\] \[x(x-8)=0\] \[x=0 \] 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

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