I'm trying to solve a rational equation and i don't know what exactly to do with the 2x^2. Here's what i have so far: (x) / (x+2) + (x) = (5x+8) / (x+2) (x+2) (x) + (x+2) (x) = (5x+8) (2x^2) + 4x = (5x+8) - 4x = - 4x 2x^2 = x+8 0 = 2x^2 + x+8 Then how do i factor this to find my x's

Question

Mathematics

Cash Simon

9 May, 08:30

I'm trying to solve a rational equation and i don't know what exactly to do with the 2x^2. Here's what i have so far: (x) / (x+2) + (x) = (5x+8) / (x+2) (x+2) (x) + (x+2) (x) = (5x+8) (2x^2) + 4x = (5x+8) - 4x = - 4x 2x^2 = x+8 0 = 2x^2 + x+8 Then how do i factor this to find my x's

+2

Answers (1)

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Brenton Leach · Answer 1

(x / (x+2)) + x = (5x+8) / (x+2)

minus (x / (x+2)) both sides and remember, a/b-c/b = (a-c) / b

x = (5x+8) / (x+2) - (x) / (x+2)

x = (5x+8-x) / (x+2)

x = (4x+8) / (x+2)

x=[ (4) (x+2) ] / (x+2)

x = (4) [ (x+2) / (x+2) ]

x=4

that was fast

I think you solved something wron since the end equation that you gave (2x^2+x+8 cannot be factore with real roots

x=4