What is 1-2x<= (x+5)^2-2(x+1) ?

The solution to 1-2x = sqrt(3)-5

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1-2x<= (x+5)^2-2(x+1)

1 - 2 x \leq (x + 5)^{2} - 2 (x + 1)

x \leq - 3 - 5 or x \geq 3 - 5

Interval Notation

(- \infty, - 3 - 5] \cup [3 - 5, \infty)

Decimal Notation

x \leq - 6.73205 \dots or x \geq - 3.26794 \dots

Solution steps

1 - 2 x \leq (x + 5)^{2} - 2 (x + 1)

Rewrite in standard form

- 10 x - 22 - x^{2} \leq 0

Complete the square

- 10 x - 22 - x^{2} : - (x + 5)^{2} + 3

- (x + 5)^{2} + 3 \leq 0

Move

3

to the right side

- (x + 5)^{2} \leq - 3

Multiply both sides by

- 1

(x + 5)^{2} \geq 3

For

u^{n} \geq a

, if

n

is even then

x + 5 \leq - 3 or x + 5 \geq 3

x + 5 \leq - 3 : x \leq - 3 - 5

x + 5 \geq 3 : x \geq 3 - 5

Combine the intervals

x \leq - 3 - 5 or x \geq 3 - 5

y^{2} - 10 y + 29 = 0

27 - w^{3}

∣ n - 9 ∣ = 14

2 t^{2} - 3 t - 14

- 9 + 3 c = 5 c + 7

What is 1-2x<= (x+5)^2-2(x+1) ?
The solution to 1-2x<= (x+5)^2-2(x+1) is x<=-sqrt(3)-5\lor x>= sqrt(3)-5