What is (x-2)^2>(x+2)-(x-2)+8 ?

The solution to (x-2)^2>(x+2)-(x-2)+8 is x 2sqrt(3)+2

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(x-2)^2>(x+2)-(x-2)+8

(x - 2)^{2} > (x + 2) - (x - 2) + 8

x < - 23 + 2 or x > 23 + 2

Interval Notation

(- \infty, - 23 + 2) \cup (23 + 2, \infty)

Decimal Notation

x < - 1.46410 \dots or x > 5.46410 \dots

Solution steps

(x - 2)^{2} > x + 2 - (x - 2) + 8

Rewrite in standard form

x^{2} - 4 x - 8 > 0

Complete the square

x^{2} - 4 x - 8 : (x - 2)^{2} - 12

(x - 2)^{2} - 12 > 0

Move

12

to the right side

(x - 2)^{2} > 12

For

u^{n} > a

, if

n

is even then

x - 2 < - 12 or x - 2 > 12

x - 2 < - 12 : x < - 23 + 2

x - 2 > 12 : x > 23 + 2

Combine the intervals

x < - 23 + 2 or x > 23 + 2

(1 + i) (1 + 2 i)

2 y - 7 \geq - 14

6 4^{\frac{1}{3}}

- 21 - 6 k \leq - 4 (2 k + 6) + 2 k

x^{2} + 4 = 40

What is (x-2)^2>(x+2)-(x-2)+8 ?
The solution to (x-2)^2>(x+2)-(x-2)+8 is x<-2sqrt(3)+2\lor x>2sqrt(3)+2