What is sqrt((6)^{x^2-x)-(6)^{|x+3|}}>= 0 ?

The solution to sqrt((6)^{x^2-x)-(6)^{|x+3|}}>= 0 is x = 3

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sqrt((6)^{x^2-x)-(6)^{|x+3|}}>= 0

(6)^{x^{2} - x} - (6)^{∣ x + 3 ∣} \geq 0

x \leq - 1 or x \geq 3

Interval Notation

(- \infty, - 1] \cup [3, \infty)

Solution steps

6^{x^{2} - x} - 6^{∣ x + 3 ∣} \geq 0

If n is even,

for all

u

True for all x

Find singularity points

Find non-negative values for radicals:

x \leq - 1 or x \geq 3

Combine the intervals

True for all x and (x \leq - 1 or x \geq 3)

Merge Overlapping Intervals

x \leq - 1 or x \geq 3

35 x^{2} - 63 x + 18 y - 10 x y

∣ 4 x + 5 ∣ \leq 10

8 x - 9 = 11 x + 12

\frac{1}{2} - \frac{3}{10}

4 m^{2} + 8 m - 5

What is sqrt((6)^{x^2-x)-(6)^{|x+3|}}>= 0 ?
The solution to sqrt((6)^{x^2-x)-(6)^{|x+3|}}>= 0 is x<=-1\lor x>= 3