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kdarnell08

  • 3 years ago

solve the square root of 6x+1 = x-1

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  1. kdarnell08
    • 3 years ago
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    |dw:1358734970247:dw|

  2. CalebBeavers
    • 3 years ago
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    Ok so first square both sides to eliminate the square root

  3. kdarnell08
    • 3 years ago
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    what do you mean square both sides...

  4. CalebBeavers
    • 3 years ago
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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 \[6x+1=(x-1)^{2}\]

  5. kdarnell08
    • 3 years ago
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    k

  6. CalebBeavers
    • 3 years ago
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    Ok so now expand the right side

  7. kdarnell08
    • 3 years ago
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    k

  8. anonymous
    • 3 years ago
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    you got this?

  9. CalebBeavers
    • 3 years ago
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    Whataya have after you expand the right side of the equation?

  10. CalebBeavers
    • 3 years ago
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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

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