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charlotteakina Group Title

Question? I need step by step instructions.

  • 2 years ago
  • 2 years ago

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  1. dpaInc Group Title
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    square both sides first... what do you get?

    • 2 years ago
  2. znimon Group Title
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    \[(\sqrt{6x-8})^2 = x^2\]

    • 2 years ago
  3. dpaInc Group Title
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    when you square a square root, you get back whatever is in the radicand...

    • 2 years ago
  4. znimon Group Title
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    The radical is over the entire left side of the equation. If you square some number then take the square root of that same number its like you did nothing to that number. In this case think of the whole left side of the equation as some number.

    • 2 years ago
  5. Jemurray3 Group Title
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    \[ \left(\sqrt{\text{SOMETHING}}\right)^2 = \text{SOMETHING} \]

    • 2 years ago
  6. dpaInc Group Title
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    \(\large (\sqrt{radicand})^2 = radicand \)

    • 2 years ago
  7. znimon Group Title
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    \[x^2 = x^2 \]

    • 2 years ago
  8. znimon Group Title
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    \[6x-8 = x^2\]

    • 2 years ago
  9. znimon Group Title
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    What should you do next?

    • 2 years ago
  10. znimon Group Title
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    You could do that but you want to get the x's together because you are solving for x.

    • 2 years ago
  11. znimon Group Title
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    Well the most efficient thing to do would be to subtract 6x from both , right?

    • 2 years ago
  12. znimon Group Title
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    So whats next?

    • 2 years ago
  13. znimon Group Title
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    6x-6x = 0 Remember "like terms"

    • 2 years ago
  14. znimon Group Title
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    \[-8 = x^2 - 6x\]

    • 2 years ago
  15. znimon Group Title
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    If we have 6x that means six times x or we have six x's. We could say our x is equal to apples. If you have six apples one would need to take six apples away to get zero. Its the same concept here to get zero and therefore get ride of 6x from the left side we need to take six x's away. And so we don't change the meaning of the equation we have to take six x's away from both sides.

    • 2 years ago
  16. dpaInc Group Title
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    still need help?

    • 2 years ago
  17. dpaInc Group Title
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    ok.... you ended up with with \(\large -8=x^2-6x \)... this is a quadratic equation and to solve quadratic equations you must first get it in standard form: \(\large 0=ax^2+bx+c \) so all you need to do is move that -8 over to the right side: \(\large 0=x^2-6x+8 \)

    • 2 years ago
  18. dpaInc Group Title
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    bye... i'll continue...

    • 2 years ago
  19. dpaInc Group Title
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    one of the easiest way to solve a quadratic equation (when possible) is to factor: \(\large 0=x^2-6x+8 \) \(\large 0=(x-4)(x-2) \) from here its obvious that x=4 or x=2

    • 2 years ago
  20. dpaInc Group Title
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    but you must now check for extraneous roots against your original problem: \(\large \sqrt{6x-8}=x \) check both solutions to see which one works....

    • 2 years ago
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