hannahmegan1
hannahmegan1 hannahmegan1
  • 10-09-2019
  • Mathematics
contestada

Part III: Determine the number and type of solutions for the following equations. Show your work.
9.) 4x2 – 28x + 49 = 0

Respuesta :

krzyzanowskijad krzyzanowskijad
  • 10-09-2019

Answer:

x=[tex]\frac{57}{28}[/tex]

Step-by-step explanation:

  • 4 x 2 – 28x + 49 = 0

multiply the numbers ( 4 x 2 )

  • 8 - 28x + 49 = 0

add the numbers ( 8+49 )

  • 8 - 28x + 49 = 0
  • 57 - 28x = 0

move the constant to the right side and change it´s sign

  • 57 - 28x = 0
  • -28x = -57

divide both sides of the equation by -28

  • -28x = -57
  • -28x \ ( -28 ) = -57 \ ( -28 )
  • x = -57 \ ( -28 )
  • x = [tex]\frac{57}{28}[/tex]

the answer would be 57 over 28 ( fraction )

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