How many integers satisfy this inequality?

First things first, I think the options here mean 4,3, 2, 1, and 0 IMHO and not -4,-3,-2,-1 and -0 (there's no meaning to -0, that's a clear clue). The hyphens could have been left out!

This is not a difficult question, conceptually. Since both the extremities of the inequality are positive numbers, we can take the square root directly.
But, because the term in the middle is a perfect square, we need to consider both positive and negative values.

\(x^2\) + 2x + 1 = \((x+1)^2\). Therefore, we can say, 4 < \((x+1)^2\) < 16. Taking the square root of all terms of the inequality,

2 < (x+1) < 4 OR -4 < (x+1) < -2.

Subtracting 1 from both sides of the inequality, we have 1 < x < 3 OR -5 < x < -3. So, there are two values that x can take i.e. x = 2 OR x = -4.

An easy way of verifying your answer is to plug these values back into the inequality given in the question statement and see if it satisfies.

The correct answer option is C.

Hope this helps!

we get 2 equations:

x^2 + 2x - 3 >0 & X^2 + 2x - 15 <0

Solving these 2 equations,

We get, X<-3, X>1 & (-5,3).

The integers that satisfy this equation are -4 & 2.

Ans C

4 < x^2 + 2x + 1 < 16
4< (x+1)^2 < 16
Taking square root

2< |x+1 | < 4

Combined solution set is (-5,-3) and (1,3). Only possible solution are -2, 2
Hence C is correct.

We need to find How many integers satisfy this inequality 4 < x^2 + 2x + 1 < 16 ?

\(2^2 < (x+1)^2 < 4^2\)

Since x is an integer
=> (x+1)^ = 3^2
=> x + 1 = +3 or -3
=> x = 2 , -4
=> 2 values

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Inequalities