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

[VJesus12]

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?

[EconomistGMATTutor]

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.

[Scott@TargetTestPrep]

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?

Factoring, we have:

(x + 10)(x + 2) < 0

We see that when x is any integer from -9 to -3, inclusive, one of the factors is positive, and the other is negative, and thus the expression is less than zero. Since there are -3 - (-9) + 1 = 7 integers from -9 to -3, inclusive, there are 7 integer solutions.

Answer: B

[Brent@GMATPrepNow]

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

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

GIVEN: x² + 12x + 20 < 0
Factor: (x + 2)(x + 10) < 0
In other words, (x + 2)(x + 10) is NEGATIVE

We can see that when x = -2 and x = -10, (x + 2)(x + 10) is ZERO
When we test some values, we can see the following:
If x < -10, then (x + 2)(x + 10) = (NEGATIVE)(NEGATIVE) = POSITIVE
If -10 < x < -2, then (x + 2)(x + 10) = (NEGATIVE)(POSITIVE) = NEGATIVE
If -2 < x, then (x + 2)(x + 10) = (POSITIVE)(POSITIVE) = POSITIVE

So, the only time (x + 2)(x + 10) is NEGATIVE is when -10 < x < -2
If x is an INTEGER, then x can equal -9, -8, -7, -6, -5, -4, and -3 (7 values)

Answer: B

Cheers,
Brent