How many integers $x$ satisfy: $-4<x^2+5x<14$?

Question

How many integers are there in the set $\{x\in\Bbb R\mid-4<x^2+5x<14\}$?

The correct answer is four, but I only find $0$ and $1$ as integers in the set.

Is the answer wrong?

Answer 1

Let's repose the constraint: $$ -4<x^2+5x<14\iff-16<4x^2+20x<56\iff 9<4x^2+20x+25<81. $$ Recognizing squares, we have $$ 3^2<(2x+5)^2<9^2\iff 2x+5=\pm 4,\pm 5,\pm 6,\pm 7,\pm 8. $$ Note you want integer $x$ so $2x+5$ is odd, leaving 4 possible values for $2x+5$: $\pm 5, \pm 7$, corresponding to $4$ values of $x$: $-6,-5,0,1$.

Answer 2

This looks like a homework problem with the following intended solution:

Consider $-4<x^2+5x<14$ one inequality at a time.

(In each case, it may be helpful to draw a graph of the corresponding parabola.)

First, $x^2 + 5x > -4$ iff $(x+1)(x+4) = x^2 + 5x + 4 > 0$.

So, we have the points $x < -4$ or $x > -1$.

Second, $x^2 + 5x < 14$ iff $(x+7)(x-2) = x^2 + 5x - 14 < 0$.

So, we have the points $-7<x<2$.

Among integers, this means considering $\{-6, -5, -4, -3, -2, -1, 0, 1\}$.

Within this set, and returning to the first inequality, which integers are less than $-4$?

Answer: $-6$ and $-5$.

Similarly, which integers are greater than $-1$?

Answer: $0$ and $1$.

Thus, the answer to your question is the four integers $\{-6, -5, 0, 1\}$.