How to resolve ?

Question

#-3x^2 +2x+5 >0#

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Answer 1 · 2018-05-08T16:37:35

#color(blue)( -1< x< 5/3)#

#-3x^2+2x+5>0#

Multiply by #-1#:

#3x^2-2x-5<0#

Factor:

#(3x-5)(x+1)<0#

Finding the boundaries:

#(3x-5)(x+1)=0=>x=5/3 and x=-1#

We know that if:

#(3x-5)(x+1)<0#

Then one bracket must be positive and the other must be negative.

i.e.

#+ - <0#

#- + <0#

So we test using our boundary values.

For

#-1< x< 5/3#

say x=1:

#-+<0# This is true.

For

#x <-1#

say #x= -2#

#- - <0# False

For

#x > 5/3#

say #x=3#

#+ + < 0# False

So the solution is:

#-1< x< 5/3#