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

The solution to x^2+6x-7<= 2 is -3sqrt(2)-3<= x<= 3sqrt(2)-3

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x^2+6x-7<= 2

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

- 32 - 3 \leq x \leq 32 - 3

Interval Notation

[- 32 - 3, 32 - 3]

Decimal Notation

- 7.24264 \dots \leq x \leq 1.24264 \dots

Solution steps

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

Rewrite in standard form

x^{2} + 6 x - 9 \leq 0

Complete the square

x^{2} + 6 x - 9 : (x + 3)^{2} - 18

(x + 3)^{2} - 18 \leq 0

Move

18

to the right side

(x + 3)^{2} \leq 18

For

u^{n} \leq a

, if

n

is even then

- 18 \leq x + 3 \leq 18

a \leq u \leq b

then

a \leq u and u \leq b

- 18 \leq x + 3 and x + 3 \leq 18

- 18 \leq x + 3 : x \geq - 32 - 3

x + 3 \leq 18 : x \leq 32 - 3

Combine the intervals

x \geq - 32 - 3 and x \leq 32 - 3

Merge Overlapping Intervals

- 32 - 3 \leq x \leq 32 - 3

x^{6} - x^{5} + x^{4} - x^{3} + x^{2} - x < 0

(2 x + 1) (10 - 3 x) < 0

(x + 1) (x - 1) < 0

\frac{1}{x - 1} \leq 2

\frac{x}{x - 2} > \frac{1}{x + 3}

What is x^2+6x-7<= 2 ?
The solution to x^2+6x-7<= 2 is -3sqrt(2)-3<= x<= 3sqrt(2)-3