What is 2x^2>= (x+1)^2 ?

The solution to 2x^2>= (x+1)^2 is x = sqrt(2)+1

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2x^2>= (x+1)^2

2 x^{2} \geq (x + 1)^{2}

x \leq - 2 + 1 or x \geq 2 + 1

Interval Notation

(- \infty, - 2 + 1] \cup [2 + 1, \infty)

Decimal Notation

x \leq - 0.41421 \dots or x \geq 2.41421 \dots

Solution steps

2 x^{2} \geq (x + 1)^{2}

Rewrite in standard form

x^{2} - 2 x - 1 \geq 0

Complete the square

x^{2} - 2 x - 1 : (x - 1)^{2} - 2

(x - 1)^{2} - 2 \geq 0

Move

2

to the right side

(x - 1)^{2} \geq 2

For

u^{n} \geq a

, if

n

is even then

x - 1 \leq - 2 or x - 1 \geq 2

x - 1 \leq - 2 : x \leq - 2 + 1

x - 1 \geq 2 : x \geq 2 + 1

Combine the intervals

x \leq - 2 + 1 or x \geq 2 + 1

\frac{( 2 x + 3 )}{( x + 1 )} > 3

\frac{( x - 3 )}{( x ^{2} + 9 )} < 1

4 x^{3} - 12 x^{2} > 0

(x + 7) (8 + 2 x - x^{2}) + (x^{2} + 3 x - 28) > 0

x + 1 \leq \frac{1}{( x - 1 )}

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