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countonme123 Group Title

Solve quadratic equation by completing the square. Check both roots. problem below

  • 2 years ago
  • 2 years ago

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  1. countonme123 Group Title
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    \[2x^2-12x +16=0\]

    • 2 years ago
  2. countonme123 Group Title
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    please explain each step I am having trouble with it

    • 2 years ago
  3. Sonita Group Title
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    2(x^2-6x+8)=0 2[(x-4)(x-2)]=0

    • 2 years ago
  4. countonme123 Group Title
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    please do explain why you do each step

    • 2 years ago
  5. NotTim Group Title
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    1st step- Factor out 2. Second step- What 2 numbers add to the end value, and multiply to the middle value?

    • 2 years ago
  6. NotTim Group Title
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    Those 2 values are then put into the brackets as you see them.

    • 2 years ago
  7. NotTim Group Title
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    And here, the signs of positive and negatives are reversed because when you solve for x individually, x-4=0 x=4. Ok?

    • 2 years ago
  8. countonme123 Group Title
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    ok but there is supposed to be two answers

    • 2 years ago
  9. countonme123 Group Title
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    4 is one of them

    • 2 years ago
  10. petewe Group Title
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    |dw:1339466532153:dw|

    • 2 years ago
  11. petewe Group Title
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    |dw:1339466562615:dw|

    • 2 years ago
  12. petewe Group Title
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    |dw:1339466604939:dw|

    • 2 years ago
  13. NotTim Group Title
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    eh, you might be might be making this more difficult than it actually is

    • 2 years ago
  14. petewe Group Title
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    I did what the question asked. Solve quadratic equation by completing the square.

    • 2 years ago
  15. NotTim Group Title
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    ay. missed that. thx.

    • 2 years ago
  16. KingGeorge Group Title
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    You made a small mistake. The -8 should also be multiplied by -2.

    • 2 years ago
  17. KingGeorge Group Title
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    Just 2 rather.

    • 2 years ago
  18. KingGeorge Group Title
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    Leaving you with\[2(x^2-6x+9-9+8)=2((x-3)^2-1)=2(x-3)^2-2\]

    • 2 years ago
  19. KingGeorge Group Title
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    Set equal to 0, and get \[2(x-3)^2-2=0\implies(x-3)^2=1\]Take the root, and solve. \[x-3=\pm1\implies x=4\quad\text{or}\quad x=2\]

    • 2 years ago
  20. countonme123 Group Title
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    ok george you are right now can you explain each step

    • 2 years ago
  21. KingGeorge Group Title
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    The hardest part is the first couple steps. You start with \[2x^2-12x+16=2(x^2-6x+8)\]The next part is where you complete the square. You take half of \(-6\), square it, and then add/subtract it. So you get\[2(x^2-6x+8)=2(x^2-6x+\left(\frac{-6}{2}\right)^2-\left(\frac{-6}{2}\right)^2+8=2(x^2-6x+9-9+8)\]

    • 2 years ago
  22. KingGeorge Group Title
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    Now you simplify as I described before. It's important that you simplify \[x^2-6x+9-9+8\]as \((x-3)^2-1\). That's the whole point of completing the square. You add/subtract just the right amount so you can have the square of a binomial. From there, you can solve for that square easily, and then just take a square root.

    • 2 years ago
  23. KingGeorge Group Title
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    Did this make sense?

    • 2 years ago
  24. countonme123 Group Title
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    yes thank you but please view my next question

    • 2 years ago
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