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anonymous

  • 5 years ago

1X^(2) + 28 - 48 = 0

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  1. anonymous
    • 5 years ago
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    1x^2-20=0 X^2=20 x=square root of 20 though i dont know if your question accepts decimals or not

  2. anonymous
    • 5 years ago
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    if the question is actually \[x ^{2} + 2x - 48\] You need factors that when added equal 2 and when multiplied equal 48 factors of 48: 1x48 2x24 4x12 6x8 Will any of those work?

  3. gw2011
    • 5 years ago
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    If the problem is x^2+28x-48, then there are no factors of -48 that will give the middle term of +28x. So, you need to use the Quadratic Formula to solve this problem. The Quadratic Formula is: x = [-b plus or minus (square root of b^2 - 4 (a)(c))]/2a Then: a=1 b=28 c=-48 x = [-28 plus or minus (square root of (28)^2 - 4(1)(-48))]/2(1) = [-28 plus or minus (square root of 784 + 192)]/2 = [-28 plus or minus (square root of 976)]/2 = [-28 plus or minus (square roots of (16)(61))/2 = [-28 plus or minus 4(square root 61)]/2 x = -14+2(square root 61) and x = -14-2(square root 61)

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