If x^2 + 12x + 20 < 0, how many integer values can x be?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 10

The OA is B.

Hello experts. Do I need to get the solutions of the quadratic equation to know how many integers can x be? Is there another way to solve it?

Hi VJesus12,
Let's take a look at your question.

$$x^2+12x+20<0$$
$$x^2+2x+10x+20<0$$
$$x\left(x+2\right)+10\left(x+2\right)<0$$
$$\left(x+2\right)\left(x+10\right)<0$$

The product can only be negative if any one of the binomial is less than zero.
There can be two cases:
Case I:
$$x+2>0,\ x+10<0$$
$$x>-2,\ x<-10$$

No value lie in this range.

Case II:
$$x+2<0,\ x+10>0$$
$$x<-2,\ x>-10$$
Therefore, values of x lie between -2 and -10 and both of end values will not include.
The values of x will be:
$$_{\left\{-3,\ -4,\ -5,\ -6,\ -7,\ -8,\ -9\right\}}$$
7 values of x.
Hence option B is correct.

Hope it helps.
I am available if you'd like any follow up.