What is (x+1)(x+2)<= 9 ?

The solution to (x+1)(x+2)<= 9 is (-sqrt(37)-3)/2 <= x<= (sqrt(37)-3)/2

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(x+1)(x+2)<= 9

(x + 1) (x + 2) \leq 9

\frac{- 37 - 3}{2} \leq x \leq \frac{37 - 3}{2}

Interval Notation

[\frac{- 37 - 3}{2}, \frac{37 - 3}{2}]

Decimal Notation

- 4.54138 \dots \leq x \leq 1.54138 \dots

Solution steps

(x + 1) (x + 2) \leq 9

Rewrite in standard form

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

Complete the square

x^{2} + 3 x - 7 : (x + \frac{3}{2})^{2} - \frac{37}{4}

(x + \frac{3}{2})^{2} - \frac{37}{4} \leq 0

Move

\frac{37}{4}

to the right side

(x + \frac{3}{2})^{2} \leq \frac{37}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{37}{4} \leq x + \frac{3}{2} \leq \frac{37}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{37}{4} \leq x + \frac{3}{2} and x + \frac{3}{2} \leq \frac{37}{4}

- \frac{37}{4} \leq x + \frac{3}{2} : x \geq \frac{- 37 - 3}{2}

x + \frac{3}{2} \leq \frac{37}{4} : x \leq \frac{37 - 3}{2}

Combine the intervals

x \geq \frac{- 37 - 3}{2} and x \leq \frac{37 - 3}{2}

Merge Overlapping Intervals

\frac{- 37 - 3}{2} \leq x \leq \frac{37 - 3}{2}

(\frac{12}{13})^{2}

- \frac{π ( - 2 π x + 1 )}{3 3} + 2 π x = 0

0^{\frac{3}{2}}

3 - \frac{2}{3}

3 x^{2} + 4 = 8 x

What is (x+1)(x+2)<= 9 ?
The solution to (x+1)(x+2)<= 9 is (-sqrt(37)-3)/2 <= x<= (sqrt(37)-3)/2