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, $x < - 6 and x > --4$