What is 7x+6<= 2x^2 ?

The solution to 7x+6 = (sqrt(97)+7)/4

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7x+6<= 2x^2

7 x + 6 \leq 2 x^{2}

x \leq \frac{- 97 + 7}{4} or x \geq \frac{97 + 7}{4}

Interval Notation

(- \infty, \frac{- 97 + 7}{4}] \cup [\frac{97 + 7}{4}, \infty)

Decimal

x \leq - 0.71221 \dots or x \geq 4.21221 \dots

Solution steps

7 x + 6 \leq 2 x^{2}

Rewrite in standard form

7 x + 6 - 2 x^{2} \leq 0

Complete the square

7 x + 6 - 2 x^{2} : - 2 (x - \frac{7}{4})^{2} + \frac{97}{8}

- 2 (x - \frac{7}{4})^{2} + \frac{97}{8} \leq 0

Move

\frac{97}{8}

to the right side

- 2 (x - \frac{7}{4})^{2} \leq - \frac{97}{8}

Multiply both sides by

- 1

2 (x - \frac{7}{4})^{2} \geq \frac{97}{8}

Divide both sides by

2

(x - \frac{7}{4})^{2} \geq \frac{97}{16}

For

u^{n} \geq a

, if

n

is even then

u \leq - n a or u \geq n a

x - \frac{7}{4} \leq - \frac{97}{16} or x - \frac{7}{4} \geq \frac{97}{16}

x - \frac{7}{4} \leq - \frac{97}{16} : x \leq \frac{- 97 + 7}{4}

x - \frac{7}{4} \geq \frac{97}{16} : x \geq \frac{97 + 7}{4}

Combine the intervals

x \leq \frac{- 97 + 7}{4} or x \geq \frac{97 + 7}{4}

54 = 54 - 9 t

x + 8 > 12

55 \cdot 515

- 4 x + 1 \leq - 3

(x + 5) (4 x - 3) = 0

What is 7x+6<= 2x^2 ?
The solution to 7x+6<= 2x^2 is x<= (-sqrt(97)+7)/4 \lor x>= (sqrt(97)+7)/4