What is -x^2+x+1<= 0 ?

The solution to -x^2+x+1 = (sqrt(5)+1)/2

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-x^2+x+1<= 0

- x^{2} + x + 1 \leq 0

x \leq \frac{- 5 + 1}{2} or x \geq \frac{5 + 1}{2}

Interval Notation

(- \infty, \frac{- 5 + 1}{2}] \cup [\frac{5 + 1}{2}, \infty)

Decimal Notation

x \leq - 0.61803 \dots or x \geq 1.61803 \dots

Solution steps

- x^{2} + x + 1 \leq 0

Complete the square

- x^{2} + x + 1 : - (x - \frac{1}{2})^{2} + \frac{5}{4}

- (x - \frac{1}{2})^{2} + \frac{5}{4} \leq 0

Move

\frac{5}{4}

to the right side

- (x - \frac{1}{2})^{2} \leq - \frac{5}{4}

Multiply both sides by

- 1

(x - \frac{1}{2})^{2} \geq \frac{5}{4}

For

u^{n} \geq a

, if

n

is even then

x - \frac{1}{2} \leq - \frac{5}{4} or x - \frac{1}{2} \geq \frac{5}{4}

x - \frac{1}{2} \leq - \frac{5}{4} : x \leq \frac{- 5 + 1}{2}

x - \frac{1}{2} \geq \frac{5}{4} : x \geq \frac{5 + 1}{2}

Combine the intervals

x \leq \frac{- 5 + 1}{2} or x \geq \frac{5 + 1}{2}

\frac{( x - 1 )}{x} - \frac{( x + 1 )}{( x - 1 )} < 0

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

\frac{1}{( 2 x - 3 )} \leq \frac{1}{( 2 x + 5 )}

\frac{( x - 4 ) ^{2}}{x} > 0

\frac{( 2 x - 7 )}{( x - 5 )} \leq 3

What is -x^2+x+1<= 0 ?
The solution to -x^2+x+1<= 0 is x<= (-sqrt(5)+1)/2 \lor x>= (sqrt(5)+1)/2