What is x^2+2x-12<= 0 ?

The solution to x^2+2x-12<= 0 is -sqrt(13)-1<= x<= sqrt(13)-1

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x^2+2x-12<= 0

x^{2} + 2 x - 12 \leq 0

- 13 - 1 \leq x \leq 13 - 1

Interval Notation

[- 13 - 1, 13 - 1]

Decimal Notation

- 4.60555 \dots \leq x \leq 2.60555 \dots

Solution steps

x^{2} + 2 x - 12 \leq 0

Complete the square

x^{2} + 2 x - 12 : (x + 1)^{2} - 13

(x + 1)^{2} - 13 \leq 0

Move

13

to the right side

(x + 1)^{2} \leq 13

For

u^{n} \leq a

, if

n

is even then

- 13 \leq x + 1 \leq 13

a \leq u \leq b

then

a \leq u and u \leq b

- 13 \leq x + 1 and x + 1 \leq 13

- 13 \leq x + 1 : x \geq - 13 - 1

x + 1 \leq 13 : x \leq 13 - 1

Combine the intervals

x \geq - 13 - 1 and x \leq 13 - 1

Merge Overlapping Intervals

- 13 - 1 \leq x \leq 13 - 1

\frac{1}{2} (6 x + 8) = 2 x + 4 + x

(x^{\frac{1}{3}})^{\frac{1}{3}}

(- \frac{4}{9})^{2}

16 > 5 x

x^{3} - 2 x^{2} - 36 x + 72

What is x^2+2x-12<= 0 ?
The solution to x^2+2x-12<= 0 is -sqrt(13)-1<= x<= sqrt(13)-1