Question

Solve the equation that models the volume of the shipping box, 8(n + 2)(n + 4) = 1,144. If you get two solutions, are they both reasonable?

Answer 1

Answer:

$n=9\\n=-15$

$n=9$ is reasonable.

$n=-15$ is not reasonable, because the volume of the shipping box can't be negative.

Step-by-step explanation:

1. Apply the Distributive property to the given equation. Then, you have:

$8n^{2}+32n+16n+64=1,144\\8n^{2}+48n-1080=0$

2. You obtain a quadratic equation, then you can use the quadratic formula to solve it:

$n=\frac{-b+/-\sqrt{b^{2}-4ac}}{2a}\\a=8\\b=48\\c=-1080$

3. Then, you obtain:

$n=\frac{-48+/-\sqrt{(48)^{2}-4(8)(-1080)}}{2(8)}\\n=9\\n=-15$

4. The volume of the shipping box can't be negative. Therefore, $n=9$ is reasonable and $n=-15$ is not reasonable.

Answer 2

Answer:

$n=-15\text{ or }n=9$

Only n=9 is reasonable.

Step-by-step explanation:

We have been given that the volume of the shipping box is: $8(n + 2)(n + 4) = 1,144$.

First of all we will distribute 8 to (n+2).

$(8n+16)(n+4) = 1,144$

Using FOIL we will get,

$8n^2+32n+16n+64= 1,144$

$8n^2+48n+64= 1,144$

$8n^2+48n+64-1,144=0$

$8n^2+48n-1080=0$

Upon dividing our equation by 8 we will get,

$n^2+6n-135=0$        

Now we will factor out our given equation by splitting the middle term.

$n^2+15n-9n-135=0$

$n(n+15)-9(n+15)=0$

$(n+15)(n-9)=0$

$(n+15)=0\text{ or }(n-9)=0$

$n=-15\text{ or }n=9$

Since the volume of a box can not be negative, therefore, only one solution is reasonable, that is n=9.