How do you solve #x^2+5x+6>0#?

2 Answers
Mar 22, 2018

Answer:

#x_1> -2 or x_2< -3#

Explanation:

#x^2+5x+6>0#
#color(blue)(x^2+2*5/2x+(5/2)^2)-(5/2)^2+6>0#
#color(blue)((x+5/2)^2)-25/4+6>0#
#(x+5/2)^2-1/4>0 |+1/4 #
#(x+5/2)^2>1/4|sqrt()#
#x+5/2>+-1/2|-5/2#

#x_1> -2 or x_2< -3#

If you struggle to understand any of the steps I made,
feel free to write a comment :)

Mar 22, 2018

Answer:

Range satisfying the given condition: #color(blue)(x<(-3) or x >(-2)#

We can also write the solution using the interval notation as:

#color(blue)((-oo,-3) uu (-2, oo)#

Explanation:

Given:

We are given the inequality:

#color(red)(x^2+5x+6>0#

#color(purple)("Step 1")#

Write this inequality as #color(green)(x^2+5x+6=0# to factorize.

Consider the quadratic expression #x^2+5x+6#

Split the middle term to factorize as shown below:

We want two numbers that multiply together to make 6, and add up to 5.

#x^2+3x+2x+6#

Factor the first two terms and the last two terms separately:

#x(x+3)+2(x+3)#

Observe that #(x+3)# is a common factor to both the terms.

Hence, we can write our factors as #color(blue)((x+3)(x+2)#

#color(purple)("Step 2")#

We will construct the sign chart:

#(i)# Compute the signs of #(x+2):#

#x+2=0#

Add #color(red)(-2# to both sides of the equation.

#x+2+color(red)((-2))=0+color(red)((-2)#

#x+cancel 2+color(red)((-cancel 2))=0+color(red)((-2)#

#x=-2#

Hence, #x+2# is ZERO for #x=-2#

Similarly, #(x+2)# is negative for #x < -2#

#(x+2)# is positive for #x > -2#

#color(purple)("Step 3")#

#(ii)# Compute the signs of #(x+3):#

#x+3# is ZERO for #x=-3#

#x+3# is negative for #x<(-3)#

#x+3# is positive for #x>(-3)#

We will summarize and create a table of values.

enter image source here

Hence,

Range satisfying the given condition: #color(blue)(x<(-3) or x >(-2)#

We can also write the solution using the interval notation as:

#color(blue)((-oo,-3) uu (-2, oo)#

An image of the graph for the inequality is available below:

enter image source here

Hope it helps.