For what values of x is the product (x+4)(x+ 6) positive?

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Answer 1 · 2016-12-16T07:48:41

#(x+6)(x+4) > 0 " if " x> -4 " or " x<-6#.

If we are excluding the product being 0 then we can say that #(x+4)(x+6) > 0#

Method 1
#x+4 = 0, x = -4#
#x+6=0, x = -6#

Through trial and error we can deduce that,
#(x+6)(x+4) > 0 " if " x> -4 " or " x<-6#.

Method 2
#=(x(x) + x(6) + 4(x) + 4(6) > 0#
#x^2 + 10x + 24 >0#

Now we can identify our abc values and solve for x using the quadratic formula.

#a = 1, b=10, c=24#
#x=(-b+- sqrt(b^2-4ac))/(2a)#
#x_1 = (-10+sqrt(10^2-4(1)(24)))/(2(1)), x_2 = (-10-sqrt(10^2-4(1)(24)))/(2(1))#

#x_1 = (-10+sqrt(100-96))/(2), x_2 = (-10-sqrt(100-96))/(2)#

#x_1 = (-10+sqrt(4))/(2), x_2 = (-10-sqrt(4))/(2)#

#x_1 = (-10+2)/(2), x_2 = (-10-2)/(2)#

#x_1 = (-8)/(2), x_2 = (-12)/(2)#

#x_1 = -4, x_2 = -6#

Through trial and error we can deduce that,
#(x+6)(x+4) > 0 " if " x> -4 " or " x<-6#.

Answer 2 · 2016-12-16T09:37:48

#x< -6# or #x> -4#

#(x+4)(x+6)# will be positive

(A) if both #(x+4)# and #(x+6)# are positive i.e. #x+4>0# and #x+6>0# i.e. #x> -4# and #x> -6#. This is possible only if #x> -4#.

or

(B) if both #(x+4)# and #(x+6)# are negative i.e. #x+4<0# and #x+6<0# i.e. #x<-4# and #x<-6#. This is possible only if #x<-6#.

Answer 3 · 2016-12-16T12:17:59

x is outside #[-6, -4]#.
The 1-D shaded x-axis illustrates this solution. The gap is out of bounds.

graph{(x+4)(x+6) > 0x^2 [-10, 10, -1, 1]}

#(x+4)(x+6)>0#. So, the factors have the same sign.

And so, #x>-4 and x > -6 to x > -4#

and

x < -4 and x < -6 to x < -6#

Thus, #x < - 6 and x > --4#