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countonme123

  • 2 years ago

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

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  1. countonme123
    • 2 years ago
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    \[2x^2-12x +16=0\]

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

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

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

  5. NotTim
    • 2 years ago
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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?

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

  7. NotTim
    • 2 years ago
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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?

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

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

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

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

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

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

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

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

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

  17. KingGeorge
    • 2 years ago
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    Just 2 rather.

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

  19. KingGeorge
    • 2 years ago
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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. \[x-3=\pm1\implies x=4\quad\text{or}\quad x=2\]

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

  21. KingGeorge
    • 2 years ago
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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\[2(x^2-6x+8)=2(x^2-6x+\left(\frac{-6}{2}\right)^2-\left(\frac{-6}{2}\right)^2+8=2(x^2-6x+9-9+8)\]

  22. KingGeorge
    • 2 years ago
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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.

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

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

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