What is (((x-3)^2))/(((x+5)(1-x)))>= 0 ?

The solution to (((x-3)^2))/(((x+5)(1-x)))>= 0 is -5<x<1\lor x=3

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(((x-3)^2))/(((x+5)(1-x)))>= 0

\frac{( ( x - 3 ) ^{2} )}{( ( x + 5 ) ( 1 - x ) )} \geq 0

- 5 < x < 1 or x = 3

Interval Notation

(- 5, 1) \cup x = 3

Solution steps

\frac{( x - 3 ) ^{2}}{( x + 5 ) ( 1 - x )} \geq 0

Identify the intervals

- 5 < x < 1 or x = 3

\frac{3}{x - 2} > 4

\frac{( ( x - 1 ) ( x - 2 ) )}{( ( x + 1 ) ( x + 2 ) )} > 0

\frac{5}{( x - 2 )} > 0

\frac{( x + ( - x ^{2} ) )}{( x ^{2} + 2 x )} \geq 0

\frac{( 2 x ^{2} + 3 x + 5 )}{( 2 x ^{2} + 8 x + 1 )} \leq 0

What is (((x-3)^2))/(((x+5)(1-x)))>= 0 ?
The solution to (((x-3)^2))/(((x+5)(1-x)))>= 0 is -5<x<1\lor x=3