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kdsim

  • 3 years ago

Math Check: Use the quadratic equation to solve: 2x^2-3x=-5 I got\frac{3\pm i \sqrt{49}}{4}\

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  1. kdsim
    • 3 years ago
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    grr\[\frac{3\pm i \sqrt{49}}{4}\ \]

  2. Brunna
    • 3 years ago
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    2x² -3x +5 = 0 delta: (-3)² - 4*5 *2 = = 9 - 40 = -31 ~ Are you sure the equation is 2x² -3x+5 =0 not 2x²-3x-5 =0 ?

  3. kdsim
    • 3 years ago
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    well it was 2x^2-3x=-5, that translates to 2x-3x+5-0 doesn't it? :o

  4. kdsim
    • 3 years ago
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    =0 not -0 sorry

  5. Brunna
    • 3 years ago
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    so, If is 2x² -3x +5=0 delta = b² -4*a*c = -31 and \[(-b \pm \sqrt{\Delta})\div 2 *a =\] \[(-(-3)\pm \sqrt{-31}) / 4 =\] \[(3\pm i \sqrt{31} )/ 4\] \[x' = (3+ i \sqrt{31}) /4\] \[x'' = (3- i \sqrt{31}) /4\]

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