What is 5x+4<= 2x^2 ?

The solution to 5x+4 = (sqrt(57)+5)/4

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5x+4<= 2x^2

5 x + 4 \leq 2 x^{2}

x \leq \frac{- 57 + 5}{4} or x \geq \frac{57 + 5}{4}

Interval Notation

(- \infty, \frac{- 57 + 5}{4}] \cup [\frac{57 + 5}{4}, \infty)

Decimal Notation

x \leq - 0.63745 \dots or x \geq 3.13745 \dots

Solution steps

5 x + 4 \leq 2 x^{2}

Rewrite in standard form

5 x + 4 - 2 x^{2} \leq 0

Complete the square

5 x + 4 - 2 x^{2} : - 2 (x - \frac{5}{4})^{2} + \frac{57}{8}

- 2 (x - \frac{5}{4})^{2} + \frac{57}{8} \leq 0

Move

\frac{57}{8}

to the right side

- 2 (x - \frac{5}{4})^{2} \leq - \frac{57}{8}

Multiply both sides by

- 1

2 (x - \frac{5}{4})^{2} \geq \frac{57}{8}

Divide both sides by

2

(x - \frac{5}{4})^{2} \geq \frac{57}{16}

For

u^{n} \geq a

, if

n

is even then

x - \frac{5}{4} \leq - \frac{57}{16} or x - \frac{5}{4} \geq \frac{57}{16}

x - \frac{5}{4} \leq - \frac{57}{16} : x \leq \frac{- 57 + 5}{4}

x - \frac{5}{4} \geq \frac{57}{16} : x \geq \frac{57 + 5}{4}

Combine the intervals

x \leq \frac{- 57 + 5}{4} or x \geq \frac{57 + 5}{4}

- 7 x - 5 = - 10 x - 26

1 1^{5}

x^{2} - 22 x + 96

- x^{2} - 1.5 x + 5 = 7

x^{2} - 3 x + 8 \geq 0

What is 5x+4<= 2x^2 ?
The solution to 5x+4<= 2x^2 is x<= (-sqrt(57)+5)/4 \lor x>= (sqrt(57)+5)/4