What is sqrt(((x-3))/((x+1)))>= ((x+2))/((x-4)) ?

The solution to sqrt(((x-3))/((x+1)))>= ((x+2))/((x-4)) is x<-1\lor 3<= x<4

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sqrt(((x-3))/((x+1)))>= ((x+2))/((x-4))

\frac{( x - 3 )}{( x + 1 )} \geq \frac{( x + 2 )}{( x - 4 )}

x < - 1 or 3 \leq x < 4

Interval Notation

(- \infty, - 1) \cup [3, 4)

Solution steps

\frac{x - 3}{x + 1} \geq \frac{x + 2}{x - 4}

f (x) \geq g (x)

then

g (x) < 0 or g (x) \geq 0

For

\frac{x + 2}{x - 4} < 0 : - 2 < x < 4

For

\frac{x + 2}{x - 4} \geq 0 : x \leq - 2

Find singularity points

Find non-negative values for radicals:

x < - 1 or x \geq 3

Combine the intervals

(x \leq - 2 or - 2 < x < 4) and (x < - 1 or x \geq 3)

Merge Overlapping Intervals

x < - 1 or 3 \leq x < 4

Verify Solutions:

x < - 1

True

, 3 \leq x < 4

True

The solution is

x < - 1 or 3 \leq x < 4

x - 3 < (15 - x)

x - 2 < x - 4

x^{2} - 3 x + 2 > 2 x - 5

x - 4 < 2

(x^{2} - 1) \cdot x^{2} - x - 2 \geq 0

What is sqrt(((x-3))/((x+1)))>= ((x+2))/((x-4)) ?
The solution to sqrt(((x-3))/((x+1)))>= ((x+2))/((x-4)) is x<-1\lor 3<= x<4