What is 3/((x^2-x+1))>1 ?

The solution to 3/((x^2-x+1))>1 is -1<x<2

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3/((x^2-x+1))>1

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

- 1 < x < 2

Interval Notation

(- 1, 2)

Solution steps

\frac{3}{x ^{2} - x + 1} > 1

Rewrite in standard form

\frac{- x ^{2} + x + 2}{x ^{2} - x + 1} > 0

Factor

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

\frac{- ( x + 1 ) ( x - 2 )}{x ^{2} - x + 1} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( x + 1 ) ( x - 2 ) ) ( - 1 )}{x ^{2} - x + 1} < 0 \cdot (- 1)

Simplify

\frac{( x + 1 ) ( x - 2 )}{x ^{2} - x + 1} < 0

Identify the intervals

- 1 < x < 2

\frac{( x + 1 )}{( x ( x ^{2} - 2 x - 3 ) )} < 0

\frac{( x ^{2} + 5 x + 6 )}{( x ^{2} - 1 )} \geq 0

\frac{( x + 5 )}{( 20 - 5 x )} \geq 0

x^{2} x > 0

x^{2} - 6 x - 16 \leq 0