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Find all real numbers $x$ that satisfy the inequality $x^2+5x<6-x^2-3x$. Express your answer in interval notation.

 Aug 16, 2023
 #1
avatar+126971 
+1

Rearrange the inequality as

 

2x^2 + 8x - 6 < 0              divide through  by  2

 

x^2 + 4x - 3  <  0

 

This  does not factor

 

Let's turn it  into  an equality

 

x^3 + 4x - 3  = 0       

 

The equation on the  left is a parabola which  turns upward....using the Quadratic Formula we  can  find the roots of this eqaution......the answer to the  problem  will lie  between these roots  (because the parbola will be  below the  x axis betwwen  these values, e.g, it will be < 0 )

 

1st  Root  =    ( -4 - sqrt [ 4^2 - (4 * 1 * -3) ] ) /  2 =      [ -4 - sqrt ( 28) ] / 2 =  [ -4 - 2sqrt (7) ] / 2 = -2 -sqrt (7)

The 2nd Root will be the conjugate of the  first root = -2 + sqrt (7)

 

Solution    x =  ( -2 - sqrt (7)   ,   -2 + sqrt (7) )

 

 

cool cool cool

 Aug 16, 2023

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