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

The solution to x (sqrt(13)+7)/6

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x<3(x^2-2x+1)

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

x < \frac{- 13 + 7}{6} or x > \frac{13 + 7}{6}

Interval Notation

(- \infty, \frac{- 13 + 7}{6}) \cup (\frac{13 + 7}{6}, \infty)

Decimal Notation

x < 0.56574 \dots or x > 1.76759 \dots

Solution steps

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

Rewrite in standard form

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

Complete the square

7 x - 3 - 3 x^{2} : - 3 (x - \frac{7}{6})^{2} + \frac{13}{12}

- 3 (x - \frac{7}{6})^{2} + \frac{13}{12} < 0

Move

\frac{13}{12}

to the right side

- 3 (x - \frac{7}{6})^{2} < - \frac{13}{12}

Multiply both sides by

- 1

3 (x - \frac{7}{6})^{2} > \frac{13}{12}

Divide both sides by

3

(x - \frac{7}{6})^{2} > \frac{13}{36}

For

u^{n} > a

, if

n

is even then

x - \frac{7}{6} < - \frac{13}{36} or x - \frac{7}{6} > \frac{13}{36}

x - \frac{7}{6} < - \frac{13}{36} : x < \frac{- 13 + 7}{6}

x - \frac{7}{6} > \frac{13}{36} : x > \frac{13 + 7}{6}

Combine the intervals

x < \frac{- 13 + 7}{6} or x > \frac{13 + 7}{6}

5 - 53 i

x^{2} + 4 x - 41 = 0

\frac{36 - x ^{2}}{x + 6} \geq 0

3 (6 x + 8) = 0

(1 + 2 i)^{3}

What is x<3(x^2-2x+1) ?
The solution to x<3(x^2-2x+1) is x<(-sqrt(13)+7)/6 \lor x>(sqrt(13)+7)/6