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Solve inequality by factoring help

  • Thread starter DeanBH
  • Start date
  • #1
82
0
x^2 + 8x + 15 > 0 solve inequality

(X+5)(X+3)>0


x+5>0 x>-5
X+3>0 x>-3

Why is the equality for X+5>0 x<-5 ? as the answers give?
 

Answers and Replies

  • #2
Because you have two possible solutions for (x+5)(x+3)>0

the first one:

x+5>0 ; x>-5

x+3>0 ; x>-3

[tex](-5, +\propto) \cap (-3, +\propto) = (-3, +\propto)[/tex]

and the second

x+5<0 ; x<-5

x+3<0 ; x<-3

[tex](-\propto, -5) \cap (-\propto, -3) = (-\propto, -5) [/tex]

The final solution:

[tex](-\propto, -5) \cup (-3, +\propto)[/tex]

Anyway, this way is much more time-consuming... So you shouldn't use it in future.. There is much easier way...
 
  • #3
82
0
Because you have two possible solutions for (x+5)(x+3)>0

the first one:

x+5>0 ; x>-5

x+3>0 ; x>-3

[tex](-5, +\propto) \cap (-3, +\propto) = (-3, +\propto)[/tex]

and the second

x+5<0 ; x<-5

x+3<0 ; x<-3

[tex](-\propto, -5) \cap (-\propto, -3) = (-\propto, -5) [/tex]

The final solution:

[tex](-\propto, -5) \cup (-3, +\propto)[/tex]

Anyway, this way is much more time-consuming... So you shouldn't use it in future.. There is much easier way...
you could have just said

"because if X > -5 then the first bracket will become negative and then (x+5)(x+3) wouldnt be > 0."


but ok. =P

btw : i just realised why
 

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