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Geometry_Hater

  • 3 years ago

For what value of k does the equation kt^2 +5t = 0 have exactly one real solution for t?

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  1. jim_thompson5910
    • 3 years ago
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    Hint: ax^2+bx + c = 0 has exactly one solution for x when b^2 - 4ac = 0

  2. estudier
    • 3 years ago
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    kt^2 +5t = t(kt +5) = 0 already has t = 0 as a solution, regardless of k

  3. NoelGreco
    • 3 years ago
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    ...but since c=0 and b=5 no can do. k=0 works

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