Question

X²-7x-30<0x²-4x+6<0​

Answer 1

Объяснение:

Решение на фотографии

Answer 2

To solve the inequalities $x^2 - 7x - 30 < 0$ and $x^2 - 4x + 6 < 0$, you can follow these steps:

1. Solve each inequality separately by finding the roots (where the expressions equal zero) and determining the sign of the expression in each interval created by these roots.

Let's start with $x^2 - 7x - 30 < 0$:

Factor the quadratic expression:

$(x - 10)(x + 3) < 0$

Now, find the critical points by setting each factor equal to zero:

$x - 10 = 0$ gives $x = 10$ $x + 3 = 0$ gives $x = -3$

These critical points divide the number line into three intervals: $(-∞, -3)$, $(-3, 10)$, and $(10, ∞)$.

Now, test a point within each interval to determine the sign of the expression:

• For $x < -3$, pick $x = -4$: $((-4 - 10)(-4 + 3)) < 0$, which simplifies to $(-14)(-1) < 0$, meaning the expression is positive in this interval.

• For $-3 < x < 10$, pick $x = 0$: $((0 - 10)(0 + 3)) < 0$, which simplifies to $(-10)(3) < 0$, meaning the expression is negative in this interval.

• For $x > 10$, pick $x = 11$: $((11 - 10)(11 + 3)) < 0$, which simplifies to $(1)(14) < 0$, meaning the expression is negative in this interval.

So, the solution to $x^2 - 7x - 30 < 0$ is $x \in (-3, 10)$.

Now, let's solve $x^2 - 4x + 6 < 0$:

To solve this quadratic inequality, you can use the quadratic formula to find its roots:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = -4$, and $c = 6$:

$x = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(6)}}{2(1)}$

$x = \frac{4 \pm \sqrt{16 - 24}}{2}$

$x = \frac{4 \pm \sqrt{-8}}{2}$

Since the discriminant ($b^2 - 4ac$) is negative, this quadratic has complex roots, and the expression $x^2 - 4x + 6$ is always positive. Therefore, there are no real solutions to $x^2 - 4x + 6 < 0$.

In summary:

1. $x^2 - 7x - 30 < 0$ has a solution: $x \in (-3, 10)$.
2. $x^2 - 4x + 6 < 0$ has no real solutions.