Question

$(x^2-x-1)(x^2-x-7)<-5$

Answer 1

$(-2, -1) \cup (2, 3)$

Answer 2

Let $y = x^2 - x$. Substituting this into the inequality, we get: $(y - 1)(y - 7) < -5$

Expand the left side: $y^2 - 7y - y + 7 < -5$ $y^2 - 8y + 7 < -5$

Add 5 to both sides: $y^2 - 8y + 7 + 5 < 0$ $y^2 - 8y + 12 < 0$

To solve this quadratic inequality, we find the roots of the corresponding quadratic equation $y^2 - 8y + 12 = 0$. Factoring the quadratic: $(y - 2)(y - 6) = 0$ The roots are $y = 2$ and $y = 6$.

Since the coefficient of $y^2$ is positive (1 > 0), the parabola opens upwards. The inequality $y^2 - 8y + 12 < 0$ holds for the values of $y$ between the roots. So, the solution for $y$ is $2 < y < 6$.

Now, substitute back $y = x^2 - x$: $2 < x^2 - x < 6$

This is a compound inequality, which can be split into two separate inequalities:

1. $x^2 - x > 2$
2. $x^2 - x < 6$

Let's solve the first inequality: $x^2 - x > 2$ $x^2 - x - 2 > 0$ Find the roots of $x^2 - x - 2 = 0$: $(x - 2)(x + 1) = 0$ The roots are $x = 2$ and $x = -1$. Since the inequality is $x^2 - x - 2 > 0$ and the parabola $x^2 - x - 2$ opens upwards, the solution is outside the roots: $x < -1$ or $x > 2$. In interval notation: $(-\infty, -1) \cup (2, \infty)$.

Now, let's solve the second inequality: $x^2 - x < 6$ $x^2 - x - 6 < 0$ Find the roots of $x^2 - x - 6 = 0$: $(x - 3)(x + 2) = 0$ The roots are $x = 3$ and $x = -2$. Since the inequality is $x^2 - x - 6 < 0$ and the parabola $x^2 - x - 6$ opens upwards, the solution is between the roots: $-2 < x < 3$. In interval notation: $(-2, 3)$.

The solution to the original inequality is the intersection of the solutions to the two inequalities: $( (-\infty, -1) \cup (2, \infty) ) \cap (-2, 3)$

We find the intersection by considering the intervals on a number line. The interval $(-2, 3)$ is the region between -2 and 3. The set $(-\infty, -1) \cup (2, \infty)$ is the region to the left of -1 or to the right of 2.

The intersection of $(-2, 3)$ and $(-\infty, -1)$ is $(-2, -1)$. The intersection of $(-2, 3)$ and $(2, \infty)$ is $(2, 3)$.

The union of these intersections is the final solution: $(-2, -1) \cup (2, 3)$.