The solution to 2x^2+x>= 7 is x = (sqrt(57)-1)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2+x>= 7

Solution

2 x^{2} + x \geq 7

x \leq \frac{- 57 - 1}{4} or x \geq \frac{57 - 1}{4}

Interval Notation

(- \infty, \frac{- 57 - 1}{4}] \cup [\frac{57 - 1}{4}, \infty)

Decimal Notation

x \leq - 2.13745 \dots or x \geq 1.63745 \dots

Solution steps

2 x^{2} + x \geq 7

Rewrite in standard form

2 x^{2} + x - 7 \geq 0

Complete the square

2 x^{2} + x - 7 : 2 (x + \frac{1}{4})^{2} - \frac{57}{8}

2 (x + \frac{1}{4})^{2} - \frac{57}{8} \geq 0

Move

\frac{57}{8}

to the right side

2 (x + \frac{1}{4})^{2} \geq \frac{57}{8}

Divide both sides by

2

(x + \frac{1}{4})^{2} \geq \frac{57}{16}

For

u^{n} \geq a

, if

n

is even then

x + \frac{1}{4} \leq - \frac{57}{16} or x + \frac{1}{4} \geq \frac{57}{16}

x + \frac{1}{4} \leq - \frac{57}{16} : x \leq \frac{- 57 - 1}{4}

x + \frac{1}{4} \geq \frac{57}{16} : x \geq \frac{57 - 1}{4}

Combine the intervals

x \leq \frac{- 57 - 1}{4} or x \geq \frac{57 - 1}{4}

5 (x - 3) = 2 x + 6

78 = 7 t + 16 t^{2}

4 x^{2} + 12 x y + 9 y^{2}

- 1 \leq \frac{x + 1}{- 2} \leq 3

f = 3^{2} - 3 - 3

What is 2x^2+x>= 7 ?
The solution to 2x^2+x>= 7 is x<= (-sqrt(57)-1)/4 \lor x>= (sqrt(57)-1)/4