Question

20 POINTS FOR THIS
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solve for |x^2-2x-8|=2x real values​

Answer 1

|x^2 - 2x - 8| =2x

take When x > 0

x^2 - 2x - 8 = 2x

x^2 - 4x - 8 = 0

x^2 - 4x + 2x - 8 = 0

(X - 4)(X +2)= 0

X = 4 and -2

when x<1

x^2 - 2x - 8 = -2x

x^2 - 8 = 0

(x - sqrt8)(x + sqrt 8)=0

x = sqrt8 and - sqrt8

Answer 2

⭐Solution ⭐

✔Given here

| x²-2x -8 | = 2x

=>| x²-4x+2x-8| = 2x

=>| x(x-4)+2(x-4)| = 2x

=> | (x-4)(x+2)| = 2x

| x-4| = 2x or, | x+2| = 2x

=> (x-2x )= 4 ,or, (x-2x)= -2

=> x = -4, or, x = 2,