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evy15

  • 3 years ago

what are the solutions of 23=2(x-3)^2+7? please help?

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  1. lopus
    • 3 years ago
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    |dw:1353177305322:dw|

  2. hartnn
    • 3 years ago
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    subtract 7 from both sides

  3. evy15
    • 3 years ago
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    I got 16=2(x-3)^2

  4. dragonsly
    • 3 years ago
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    |dw:1353177800615:dw|

  5. anas2000
    • 3 years ago
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    3± 2√2

  6. anas2000
    • 3 years ago
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    First square (x-3). You will need to FOIL this. (x-3)(x-3) = x^2-3x-3x+9 = x^2-6x+9 Then, multiply by 2. 23 = 2(x^2-6x+9)+7 = (2x^2-12x+18)+7 Add 7 to 18. 23 = 2x^2-12x+25 Then subtract 23 on both sides. 0 = 2x^2-12x+2 The right side all have the common factor of 2, so divide both sides by 2. 0 = x^2-6x+1 If you try to factor the equation, you'll notice that it won't work, so you'll need to use the quadratic equation. -b ± √(b^2 - 4ac) / 2a x^2 (1) = a -6x (-6) = b 1 = c Plug in the numbers. 6± √[(-6)^2 - 4(1)(1)] / 2(1) 6± √(36-4) / 2 6± √32 / 2 Simplify. 6± √16*2 / 2 6± 4√2 / 2 = 3± 2√2

  7. evy15
    • 3 years ago
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    ok I see, thanks everyone!! (:

  8. anas2000
    • 3 years ago
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    your welcome

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