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Jeans123

  • 3 years ago

Solve the quadratic equation by completing the square x^2-12+7=0

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  1. Jeans123
    • 3 years ago
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    @cruffo

  2. phi
    • 3 years ago
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    if you have time, this might help http://www.khanacademy.org/math/algebra/quadtratics/v/completing-the-square-2

  3. cruffo
    • 3 years ago
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    Humm. Quadratic... factor? complete the square? or quadratic formula?

  4. cruffo
    • 3 years ago
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    Ah ha!! instructions say "completing the square", so I guess we'll completing the square :)

  5. cruffo
    • 3 years ago
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    BTW... was the problem x^2-12x+7=0 (original was missing the x on the 12)

  6. Jeans123
    • 3 years ago
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    oops yes:)

  7. cruffo
    • 3 years ago
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    cool. x^2 -12x + 7 = 0 step 1: move the constant over x^2-12x= -7

  8. cruffo
    • 3 years ago
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    step 2: take half the bx term \(12 \div 2 = 6\)

  9. cruffo
    • 3 years ago
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    step 3: square the number in step # \(6^2 = 36\)

  10. cruffo
    • 3 years ago
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    step 4: add the number from step 3 to both sides: \[x^2 + 12x + 36 = -7 + 36\]

  11. cruffo
    • 3 years ago
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    so far so good?

  12. Jeans123
    • 3 years ago
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    Yep:)

  13. cruffo
    • 3 years ago
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    \[x^2 + 12x + 7 = 0\] \[x^2 + 12x = -7\] \[x^2 + 12x + 36 = -7+36\] step 5: factor the left-hand side, and simplify the right-hand side: \[(x+6)^2 = 29\] Hint - the left-hand side always factors to (x + # from step 2, including the sign, either + or -)^2

  14. cruffo
    • 3 years ago
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    from this point, use the square root property to finish solving for x.

  15. Jeans123
    • 3 years ago
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    x^2+12x+7=0?

  16. Jeans123
    • 3 years ago
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    that is what I started with?

  17. cruffo
    • 3 years ago
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    did you just FOIL (x+6)^2 and subtract 29 to zero out the equation! Yep, that is back to where you started... :)

  18. cruffo
    • 3 years ago
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    "Square Root Property" means take the square root \(\sqrt{\;\;\;}\) of both sides .

  19. cruffo
    • 3 years ago
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    \[\sqrt{(x+6)^2} = \pm \sqrt {29}\]

  20. Jeans123
    • 3 years ago
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    Ohh!!!:D

  21. cruffo
    • 3 years ago
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    :)

  22. Jeans123
    • 3 years ago
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    I got x=+- -.6148

  23. cruffo
    • 3 years ago
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    \[\sqrt{(x+6)^2} = \pm \sqrt {29}\] humm.. \(\sqrt{29}\) does not simplify (other than decimal approx) so I'm gonna leave it for now... \[x+6 = \pm \sqrt{29}\] subtract 6 from both sides (but not from 29, that's inside the square root) \[x = -6 \pm \sqrt{29}\] that gives us two solutions: \[x = -6 + \sqrt{29} \approx -0.615\] and \[x = -6 - \sqrt{29} \approx -11.385 \]

  24. Jeans123
    • 3 years ago
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    Thank you sooo much!!!!!:D

  25. cruffo
    • 3 years ago
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    np :)

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