Respuesta :

sqdancefan

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Answer:

  k > 9/20

Step-by-step explanation:

If the parabola does not touch the x-axis, the discriminant is negative. That means ...

  b^2 -4ac < 0

  (-3)^2 -4(5)(k) < 0

  9 < 20k . . . . . . . . . add 20k

  k > 9/20 . . . . . . . . .divide by 20

VectorFundament120

[tex]y=5x^2-3x+k\\[/tex]

Discriminant  |  D < 0 (Because doesn't intercept any x-values)

[tex]D = (-3)^2-4(5)(k) < 0\\9-20k<0\\-20k<-9\\k>\frac{9}{20}[/tex]

Therefore, the value of k must be greater than 9/20

so k > 9/20

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