Question

(x²-x-1)(x²-x-7)≤-5 У відповідь записати добуток усіх цілих розвязків

Answer 1

Решение на фотографии

Answer 2

To find the product of all integer solutions for the inequality $(x^2-x-1)(x^2-x-7) \leq -5$, we first need to determine the values of $x$ that satisfy the inequality.

Step 1: Solve each quadratic equation:

1. $x^2 - x - 1 = 0$: We can use the quadratic formula to find the roots: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -1$, and $c = -1$.

$x = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{1 \pm \sqrt{5}}{2}$

So the roots are $x = \frac{1 + \sqrt{5}}{2}$ and $x = \frac{1 - \sqrt{5}}{2}$.

2. $x^2 - x - 7 = 0$: Again, we use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = -1$, and $c = -7$.

$x = \frac{1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} = \frac{1 \pm \sqrt{29}}{2}$

So the roots are $x = \frac{1 + \sqrt{29}}{2}$ and $x = \frac{1 - \sqrt{29}}{2}$.

Step 2: Now that we have the roots, we can plot them on the number line and find the intervals where the inequality holds true.

Number line with roots:

$$\begin{array}{ccccccccccccc} \ldots & \frac{1 - \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} & \frac{1 - \sqrt{29}}{2} & \frac{1 + \sqrt{29}}{2} & \ldots \\ \end{array}$$

Step 3: Test intervals to find the valid solution regions:

We can pick test points from each interval and check if the inequality holds true:

1. Test $x = 0$: $(0^2 - 0 - 1)(0^2 - 0 - 7) = (-1)(-7) = 7 > -5$ -> False

2. Test $x = 1$: $(1^2 - 1 - 1)(1^2 - 1$