asar49 asar49
  • 15-08-2021
  • Mathematics
contestada

Find k such that 16x2 – 8 /3x+k=0 has a repeated real solution.

Respuesta :

jimrgrant1 jimrgrant1
  • 15-08-2021

Answer:

k = [tex]\frac{1}{9}[/tex]

Step-by-step explanation:

Using the discriminant

Δ = b² - 4ac

For a repeated real solution then

b² - 4ac = 0

Given

16x² - [tex]\frac{8}{3}[/tex] x + k = 0

with a = 16, b = - [tex]\frac{8}{3}[/tex] , c = k

(- [tex]\frac{8}{3}[/tex] )² - ( 4 × 16 × k) = 0

[tex]\frac{64}{9}[/tex] - 64k = 0 ( subtract [tex]\frac{64}{9}[/tex] from both sides )

- 64k = - [tex]\frac{64}{9}[/tex] ( divide both sides by - 64 )

k = [tex]\frac{1}{9}[/tex]

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