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The real number  satisfies \($x^2 - 5x + 6 < 0.$\) Find all possible values of \($x^2 + 5x + 6.$\)

 Jun 25, 2021
 #1
avatar+151 
+3

we are given $  x^2+5x+6<0 $ right? turn the $<$ into $=$:

 

$  x^2+5x+6=0 $

 

again, a binomial in which $  a=1$   ;   $b=5$   ;   $c=6 $

 

plugging that into $ _1x_2=\frac{-b\pm \sqrt{b^2-4ac}}{2a}  $ we get:

 

$ _1x_2= \frac{-5\pm \sqrt{5^2-4\cdot \:1\cdot \:6}}{2\cdot \:1}  $

 

$  _1x_2=  \frac{-5\pm \:\sqrt{25-24} }{2\cdot \:1} $

 

$   _1x_2= \frac{-5\pm \:1}{2\cdot \:1} $

that gives us:

$ x_1=-2$   and  $x=-3 $

 Jun 25, 2021
 #2
avatar+121048 
+2

x^2 +  5x +  6 <   0

 

(x + 3) ( x + 2)  <  0

 

This wil be true  when      -3 < x < -2 

 

 

cool cool cool

 Jun 25, 2021
 #3
avatar+151 
+3

i did not send through :/

UsernameTooShort  Jun 25, 2021

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