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

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x^2+x<= 1

Solution

x^{2} + x \leq 1

\frac{- 5 - 1}{2} \leq x \leq \frac{5 - 1}{2}

Interval Notation

[\frac{- 5 - 1}{2}, \frac{5 - 1}{2}]

Decimal Notation

- 1.61803 \dots \leq x \leq 0.61803 \dots

Solution steps

x^{2} + x \leq 1

Rewrite in standard form

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}

For

u^{n} \leq a

, if

n

is even then

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

a \leq u \leq b

then

a \leq u and u \leq b

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

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

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

Combine the intervals

x \geq \frac{- 5 - 1}{2} and x \leq \frac{5 - 1}{2}

Merge Overlapping Intervals

\frac{- 5 - 1}{2} \leq x \leq \frac{5 - 1}{2}

r^{2} - 4 r + 13 = 0

2 x^{2} - 24 x + 216 = 0

6 n^{2} + n - 2

- \frac{3}{- 4}

b^{8} c^{11} + b^{5} c^{6}

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