What is 7x+4<= 3x^2 ?

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

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7x+4<= 3x^2

7 x + 4 \leq 3 x^{2}

x \leq \frac{- 97 + 7}{6} or x \geq \frac{97 + 7}{6}

Interval Notation

(- \infty, \frac{- 97 + 7}{6}] \cup [\frac{97 + 7}{6}, \infty)

Decimal Notation

x \leq - 0.47480 \dots or x \geq 2.80814 \dots

Solution steps

7 x + 4 \leq 3 x^{2}

Rewrite in standard form

7 x + 4 - 3 x^{2} \leq 0

Complete the square

7 x + 4 - 3 x^{2} : - 3 (x - \frac{7}{6})^{2} + \frac{97}{12}

- 3 (x - \frac{7}{6})^{2} + \frac{97}{12} \leq 0

Move

\frac{97}{12}

to the right side

- 3 (x - \frac{7}{6})^{2} \leq - \frac{97}{12}

Multiply both sides by

- 1

3 (x - \frac{7}{6})^{2} \geq \frac{97}{12}

Divide both sides by

3

(x - \frac{7}{6})^{2} \geq \frac{97}{36}

For

u^{n} \geq a

, if

n

is even then

x - \frac{7}{6} \leq - \frac{97}{36} or x - \frac{7}{6} \geq \frac{97}{36}

x - \frac{7}{6} \leq - \frac{97}{36} : x \leq \frac{- 97 + 7}{6}

x - \frac{7}{6} \geq \frac{97}{36} : x \geq \frac{97 + 7}{6}

Combine the intervals

x \leq \frac{- 97 + 7}{6} or x \geq \frac{97 + 7}{6}

- 2 - 3

x^{2} - 16 x < 0

3 x^{2} + 14 x + 1 = 0

\frac{1}{3} ∣ 3 x - 1 ∣ - 3 \leq 5

\frac{1}{11} x^{2} - 11 = 0

What is 7x+4<= 3x^2 ?
The solution to 7x+4<= 3x^2 is x<= (-sqrt(97)+7)/6 \lor x>= (sqrt(97)+7)/6