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Mrbonez321

  • 2 years ago

Solve for x: 2x^2 - 4x - 14 = 0 I have everything up to 2(x-2)^2=6, but I do not know how to continue.

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  1. Chlorophyll
    • 2 years ago
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    2(x^2 - 2x - 7) = 0 Δ = 4 - ( -28) = 32 x = (-4 ± 4 √2)/ 4 = -1 ± √2

  2. Chlorophyll
    • 2 years ago
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    Oops, x = (-4 ± 4 √2)/ 2 = -2 ± 2√2

  3. Mrbonez321
    • 2 years ago
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    I dont understand Δ... I am only in Algebra II

  4. S
    • 2 years ago
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    2x^2 - 4x - 14 = 0 divide everything by 2 in order to get x^2 by itself without the coefficient x^2 - 2x - 7 = 0 and then use the quadratic formula (-b+-sqrt(b^2 - 4ac))/2a (2 +or- sqrt(4 - 4*1*-7))/2 (2 + sqrt(32)) / 2 and (2 - sqrt(32))/2 2 + 4sqrt(2) / 2 and 2 - 4sqrt(2) / 2 1 + 2sqrt(2) and 1 - 2sqrt(2)

  5. Chlorophyll
    • 2 years ago
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    That's quadratic formula in algebra II

  6. Chlorophyll
    • 2 years ago
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    Only quadratic formula can solve any cases!

  7. Mrbonez321
    • 2 years ago
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    Thank You Both!

  8. Chlorophyll
    • 2 years ago
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    S's result is perfect, not typo as mine!

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