We have #sqrt(2x+10)<3x-5#
We know #sqrt(2x+10) # is real for #sqrt(2x+10)>=0#
#sqrt(2x+10)>=0#
Square both sides.
#2x+10>=0#
Subtract #10# from both sides.
#2x+10color(red)(-10)>=0color
(red)(-10)#
#2x>=-10#
Divide by #2# on both sides
#2x/color(red)2>=-10/color(red)2#
#color(green)1.# #x>=-5#
We now have our first condition.
Now similarly solve #3x-5>0#
We will get #x>5/3# which is #x>1.6#
#color(green)2.##x>5/3#
This is our second condition.
Now we have the inequality #sqrt(2x+10)<3x-5#
Square both sides.
#2x+10<(3x-5)^2#
Use #(a-b)^2=a^2+b^2-2ab# on the LHS.
#2x+10<9x^2+25- 2*3x*5#
#2x+10<9x^2-30x+25#
Now subtract #(2x+10)# from both sides.
#2x+10color(red)(-(2x+10))<9x^2-30x+25color(red)(-(2x+10))#
#0<9x^2-32x+15#
#9x^2-32x+15>0#
Factorize #9x^2-32x+15# and we will get #(x-3)(x-5/9)#
Therefore,
#(x-3)(x-5/9)>0#
So, #(x-3)>0# and #(x-5/9)>0#
#(x>3)# and #(x>5/9)#
Remember we made two conditions during the start, we'll apply them here.
So we had #(x>=-5)# and #(x>5/3)# as the two conditions.
When we apply the two conditions to #x>3# and #x>5/9# we see that #x>5/9# is not valid.
Therefore #x>3#
