Since √(3x+1) and √(x-5) is always positive, you can square out both sides.
[tex] \sqrt{3x + 1} > \sqrt{x - 5} \\ {( \sqrt{3x + 1}) }^{2} > {( \sqrt{x - 5} )}^{2} \\ 3x + 1 > x - 5 \\ 2x > - 6 \\ x > - 3[/tex]
Since substituting x = -2, -1 , ... , 4 into √(3x+1) and √(x-5) will result into complex solutions,
Therefore,
[tex]x > 4[/tex]
Answer:
-3<x≤5
Step-by-step explanation:
√(3x+1)>√(x-5)
Square each one of the square roots. Then create 3 different inequality expressions and solve them/
1. 3x+1>x-5= x>-3
2. 3x+1≥0= x≥-1/3
3. x-5≥0= x≥5
From here I'm going to take the values that cover the most of the area in this expression which should be -3<x≤5.
