What is x^2-7x^2+14x-8<0 ?

The solution to x^2-7x^2+14x-8 4/3

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x^2-7x^2+14x-8<0

x^{2} - 7 x^{2} + 14 x - 8 < 0

x < 1 or x > \frac{4}{3}

Interval Notation

(- \infty, 1) \cup (\frac{4}{3}, \infty)

Decimal Notation

x < 1 or x > 1.33333 \dots

Solution steps

x^{2} - 7 x^{2} + 14 x - 8 < 0

Rewrite in standard form

- 3 x^{2} + 7 x - 4 < 0

Factor

- 3 x^{2} + 7 x - 4 : - (x - 1) (3 x - 4)

- (x - 1) (3 x - 4) < 0

Multiply both sides by

- 1

(reverse the inequality)

(- (x - 1) (3 x - 4)) (- 1) > 0 \cdot (- 1)

Simplify

(x - 1) (3 x - 4) > 0

Identify the intervals

x < 1 or x > \frac{4}{3}

x - 1 < \frac{1}{( x - 3 )}

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

x^{2} > - 4

x^{2} > - 2

48 + 2 x - x^{2} \geq 0