What is 600x^1+900x^2>= 1800 ?

The solution to 600x^1+900x^2>= 1800 is x = (sqrt(19)-1)/3

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600x^1+900x^2>= 1800

600 x^{1} + 900 x^{2} \geq 1800

x \leq \frac{- 19 - 1}{3} or x \geq \frac{19 - 1}{3}

Interval Notation

(- \infty, \frac{- 19 - 1}{3}] \cup [\frac{19 - 1}{3}, \infty)

Decimal Notation

x \leq - 1.78629 \dots or x \geq 1.11963 \dots

Solution steps

600 x^{1} + 900 x^{2} \geq 1800

Rewrite in standard form

3 x^{2} + 2 x - 6 \geq 0

Complete the square

3 x^{2} + 2 x - 6 : 3 (x + \frac{1}{3})^{2} - \frac{19}{3}

3 (x + \frac{1}{3})^{2} - \frac{19}{3} \geq 0

Move

\frac{19}{3}

to the right side

3 (x + \frac{1}{3})^{2} \geq \frac{19}{3}

Divide both sides by

3

(x + \frac{1}{3})^{2} \geq \frac{19}{9}

For

u^{n} \geq a

, if

n

is even then

x + \frac{1}{3} \leq - \frac{19}{9} or x + \frac{1}{3} \geq \frac{19}{9}

x + \frac{1}{3} \leq - \frac{19}{9} : x \leq \frac{- 19 - 1}{3}

x + \frac{1}{3} \geq \frac{19}{9} : x \geq \frac{19 - 1}{3}

Combine the intervals

x \leq \frac{- 19 - 1}{3} or x \geq \frac{19 - 1}{3}

3 n + 17 = 44

21 x^{3} - 18 x^{2} y + 24 x y^{2}

4. 2^{2} + 3^{2}

\frac{x}{5} + 6 < 2

c = 180 (2) + 250

What is 600x^1+900x^2>= 1800 ?
The solution to 600x^1+900x^2>= 1800 is x<= (-sqrt(19)-1)/3 \lor x>= (sqrt(19)-1)/3