What is 1 2/3 x+(2x+1)(13x-1)>= 0 ?

The solution to 1 2/3 x+(2x+1)(13x-1)>= 0 is x = (sqrt(595)-19)/(78)

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1 2/3 x+(2x+1)(13x-1)>= 0

1 \frac{2}{3} x + (2 x + 1) (13 x - 1) \geq 0

x \leq \frac{- 595 - 19}{78} or x \geq \frac{595 - 19}{78}

Interval Notation

(- \infty, \frac{- 595 - 19}{78}] \cup [\frac{595 - 19}{78}, \infty)

Decimal Notation

x \leq - 0.55631 \dots or x \geq 0.06913 \dots

Solution steps

1 \frac{2}{3} x + (2 x + 1) (13 x - 1) \geq 0

Convert mixed numbers to improper fractions:

1 \frac{2}{3} = \frac{5}{3}

\frac{5}{3} x + (2 x + 1) (13 x - 1) \geq 0

Rewrite in standard form

78 x^{2} + 38 x - 3 \geq 0

Complete the square

78 x^{2} + 38 x - 3 : 78 (x + \frac{19}{78})^{2} - \frac{595}{78}

78 (x + \frac{19}{78})^{2} - \frac{595}{78} \geq 0

Move

\frac{595}{78}

to the right side

78 (x + \frac{19}{78})^{2} \geq \frac{595}{78}

Divide both sides by

78

(x + \frac{19}{78})^{2} \geq \frac{595}{6084}

For

u^{n} \geq a

, if

n

is even then

x + \frac{19}{78} \leq - \frac{595}{6084} or x + \frac{19}{78} \geq \frac{595}{6084}

x + \frac{19}{78} \leq - \frac{595}{6084} : x \leq \frac{- 595 - 19}{78}

x + \frac{19}{78} \geq \frac{595}{6084} : x \geq \frac{595 - 19}{78}

Combine the intervals

x \leq \frac{- 595 - 19}{78} or x \geq \frac{595 - 19}{78}

6^{2}

2 (x + 10) = x + 24

2 x > 14

4 x^{4} - 20 x^{2} + 16

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

What is 1 2/3 x+(2x+1)(13x-1)>= 0 ?
The solution to 1 2/3 x+(2x+1)(13x-1)>= 0 is x<= (-sqrt(595)-19)/(78)\lor x>= (sqrt(595)-19)/(78)