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Popular >

-x^2+16<= 0

Solution

- x^{2} + 16 \leq 0

Solution

x \leq - 4 or x \geq 4

+1

Interval Notation

(- \infty, - 4] \cup [4, \infty)

Solution steps

- x^{2} + 16 \leq 0

Move

16

to the right side

- x^{2} \leq - 16

Multiply both sides by

- 1

x^{2} \geq 16

For

u^{n} \geq a

, if

n

is even then

x \leq - 16 or x \geq 16

16 = 4

x \leq - 4 or x \geq 4

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\frac{( 3 + 3 i )}{( 5 + 4 i )}

5= 1/7 (10x+28)

5 = \frac{1}{7} (10 x + 28)

\frac{x - 5}{8} = - 2

Frequently Asked Questions (FAQ)

What is -x^2+16<= 0 ?
The solution to -x^2+16<= 0 is x<=-4\lor x>= 4