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mathtard

  • 4 years ago

Solve by completing the square x^2-14x=10

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  1. LagrangeSon678
    • 4 years ago
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    take half the coefficent of -14, it should be -7. The we have (x-7)^2 which equals=x^2-14x+49. That means you get: (x-7)^2=59

  2. LagrangeSon678
    • 4 years ago
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    or (x-7)^2-59=0

  3. jim_thompson5910
    • 4 years ago
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    \[\large x^2-14x=10\] \[\large x^2-14x-10=0\] \[\large x^2-14x+49-49-10=0\] \[\large (x-7)^2-49-10=0\] \[\large (x-7)^2-59=0\] \[\large (x-7)^2=59\] \[\large x-7=\pm\sqrt{59}\] \[\large x=7\pm\sqrt{59}\]

  4. LagrangeSon678
    • 4 years ago
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    x=7plus or mius sqrt59

  5. anonymous
    • 4 years ago
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    \[x^2-14x=10\] \[(x-7)^2=10+7^2=59\] \[x-7=\pm\sqrt{59}\] \[x=7\pm\sqrt{59}\]

  6. nilankshi
    • 4 years ago
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    good

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