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help! see photo
Some of the steps for completing the square to solve x2 + 5x = 2 are shown.
x2 + 5x = 2 x2 + 5x + (StartFraction 5 Over 2 EndFraction) squared = 2 + (StartFraction 5 Over 2 EndFraction) squared (x + StartFraction 5 Over 2 EndFraction) squared = StartFraction 33 Over 4 EndFraction
Which are solutions of x2 + 5x = 2? Check all that apply
StartFraction 5 Over 2 EndFraction + StartFraction StartRoot 33 EndRoot Over 4 EndFraction StartFraction 5 Over 2 EndFraction + StartFraction StartRoot 33 EndRoot Over 2 EndFraction StartFraction 5 Over 2 EndFraction minus StartFraction StartRoot 33 EndRoot Over 2 EndFraction StartFraction negative 5 Over 2 EndFraction minus StartFraction StartRoot 33 EndRoot Over 2 EndFraction StartFraction negative 5 Over 2 EndFraction + StartFraction StartRoot 33 EndRoot Over 2 EndFraction

help see photo Some of the steps for completing the square to solve x2 5x 2 are shown x2 5x 2 x2 5x StartFraction 5 Over 2 EndFraction squared 2 StartFraction 5 class=

Respuesta :

mysticchacha

Answer:

x = -5/2 Β + root(33)/2 Β  or Β  x = -5/2 - root(33)/2

Step-by-step explanation:

ok so your computer is

saying:

x^2 Β + 5x = 2

and takes the 5 divides it by 2 Β and squares the result

x^2 Β + 5x Β + Β 25/4 Β  Β = Β  2 + 25 /4 Β  but then Β  replaces a trinomial with a binomial

(x + 5/2)^2 Β = Β 2 + 25/4 = Β 33/4

solving

x = -5/2 Β + root(33/4) Β  or Β  x = -5/2 - root(33/4)

or

x = -5/2 Β + root(33)/2 Β  or Β  x = -5/2 - root(33)/2

Answer:

x=-5/2

Step-by-step explanation: