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Find any domain restrictions on the given rational equation
х/x+4 + 12/x^2 +5x+4 = 8x/5x-15
Select all that apply.
O A. x= -1
B. x = 0
C. x= 3
D. x = -4

Sagot :

fieryanswererft

Answer:

x ≠ -4, -1, 3

Step-by-step explanation:

[tex]\frac{x}{x+4}+ \frac{12}{x^2} +5x+4 = \frac{8x}{5x-15}[/tex]

[tex]12/(x^2+5x+4) = 8x/(5x-15)[/tex]

[tex]\frac{12}{((x+1)(x+4))} = \frac{8x}{(5(x-3))}[/tex]

[tex]\\ \tt\hookrightarrow \dfrac{x}{x+4}+\dfrac{12}{x^2+5x+4}=\dfrac{8x}{5x-15}[/tex]

[tex]\\ \tt\hookrightarrow \dfrac{x}{x+4}+\dfrac{12}{(x+1)(x+4)}=\dfrac{8x}{5(x-3)}[/tex]

  • For x=-4 First fraction becomes undefined
  • For x=-1 second fraction becomes undefined
  • For x=3 third fraction also becomes undefined

So Domain=(-4,-1,3)

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