What is (x^2-7x+12)/(2x^2+4x+5)> 1/2 ?

The solution to (x^2-7x+12)/(2x^2+4x+5)> 1/2 is x< 19/18

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(x^2-7x+12)/(2x^2+4x+5)> 1/2

\frac{x ^{2} - 7 x + 12}{2 x ^{2} + 4 x + 5} > \frac{1}{2}

x < \frac{19}{18}

Interval Notation

(- \infty, \frac{19}{18})

Decimal Notation

x < 1.05555 \dots

Solution steps

\frac{x ^{2} - 7 x + 12}{2 x ^{2} + 4 x + 5} > \frac{1}{2}

Rewrite in standard form

\frac{- 18 x + 19}{2 x ^{2} + 4 x + 5} > 0

Identify the intervals

x < \frac{19}{18}

\frac{4}{( x - 3 )} < 0

\frac{( ( x - 1 ) ^{2} ( x + 1 ) ^{3} )}{( x ^{4} ( x - 2 ) )} \leq 0

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

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

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

What is (x^2-7x+12)/(2x^2+4x+5)> 1/2 ?
The solution to (x^2-7x+12)/(2x^2+4x+5)> 1/2 is x< 19/18