What is ((x^2-3x+2))/((x^2-3x+3))<4 ?

The solution to ((x^2-3x+2))/((x^2-3x+3))<4 is True for all x

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((x^2-3x+2))/((x^2-3x+3))<4

\frac{( x ^{2} - 3 x + 2 )}{( x ^{2} - 3 x + 3 )} < 4

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

\frac{x ^{2} - 3 x + 2}{x ^{2} - 3 x + 3} < 4

Rewrite in standard form

\frac{- 3 x ^{2} + 9 x - 10}{x ^{2} - 3 x + 3} < 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( 3 x ^{2} - 9 x + 10 ) ) ( - 1 )}{x ^{2} - 3 x + 3} > 0 \cdot (- 1)

Simplify

\frac{3 x ^{2} - 9 x + 10}{x ^{2} - 3 x + 3} > 0

Identify the intervals

True for all x

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

(a + 3) (a - 2) < 0

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

x^{2} \geq - 5

x^{2} + 2 x < 15

What is ((x^2-3x+2))/((x^2-3x+3))<4 ?
The solution to ((x^2-3x+2))/((x^2-3x+3))<4 is True for all x