hailey9325
hailey9325 hailey9325
  • 16-01-2020
  • Mathematics
contestada

what value(s) of b will cause 2x^2+bx+128=0 to have one real solution?

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sqdancefan
sqdancefan sqdancefan
  • 16-01-2020

Answer:

  b = -32 or +32

Step-by-step explanation:

The discriminant of the equation is ...

  d = b² -4ac

  d = b² -4(2)(128)

  d = b² -1024

In order for there to be one real solution, we must have d=0. So ...

  0 = b² -1024

  1024 = b²

  ±√1024 = ±32 = b

The value of b can be -32 or +32 to cause the equation to have one real solution.

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