What is (x+3)(2x+5)<= (2x+1)^2+6x ?

The solution to (x+3)(2x+5) = (sqrt(113)+1)/4

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(x+3)(2x+5)<= (2x+1)^2+6x

(x + 3) (2 x + 5) \leq (2 x + 1)^{2} + 6 x

x \leq \frac{- 113 + 1}{4} or x \geq \frac{113 + 1}{4}

Interval Notation

(- \infty, \frac{- 113 + 1}{4}] \cup [\frac{113 + 1}{4}, \infty)

Decimal Notation

x \leq - 2.40753 \dots or x \geq 2.90753 \dots

Solution steps

(x + 3) (2 x + 5) \leq (2 x + 1)^{2} + 6 x

Rewrite in standard form

- 2 x^{2} + x + 14 \leq 0

Complete the square

- 2 x^{2} + x + 14 : - 2 (x - \frac{1}{4})^{2} + \frac{113}{8}

- 2 (x - \frac{1}{4})^{2} + \frac{113}{8} \leq 0

Move

\frac{113}{8}

to the right side

- 2 (x - \frac{1}{4})^{2} \leq - \frac{113}{8}

Multiply both sides by

- 1

2 (x - \frac{1}{4})^{2} \geq \frac{113}{8}

Divide both sides by

2

(x - \frac{1}{4})^{2} \geq \frac{113}{16}

For

u^{n} \geq a

, if

n

is even then

x - \frac{1}{4} \leq - \frac{113}{16} or x - \frac{1}{4} \geq \frac{113}{16}

x - \frac{1}{4} \leq - \frac{113}{16} : x \leq \frac{- 113 + 1}{4}

x - \frac{1}{4} \geq \frac{113}{16} : x \geq \frac{113 + 1}{4}

Combine the intervals

x \leq \frac{- 113 + 1}{4} or x \geq \frac{113 + 1}{4}

2 x^{3} + 4 x^{2} - 13 x + 7 < 0

x^{3} \geq x^{2}

\frac{( x ^{2} + 4 x + 4 )}{( 2 x ^{2} - x - 1 )} < 0

(\frac{5}{3}) < \frac{( 2 x + 1 )}{x}

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

What is (x+3)(2x+5)<= (2x+1)^2+6x ?
The solution to (x+3)(2x+5)<= (2x+1)^2+6x is x<= (-sqrt(113)+1)/4 \lor x>= (sqrt(113)+1)/4