Question

X2-x+6>0(x-5)(x+5)<0​

Answer 1

Ответ:

х=1

1×2-1+6=7

7>0

(1-5)(1+5)=-4×6=-24

-24<0

Answer 2

1)X^2 - x - 6 = 0

D = 1 + 4*6 = 25

x1 = ( 1 + 5)/2 = 6/2 = 3;

x2 = ( 1 - 5)/2 = - 4/2 = - 2;

2) 2/5 x=0

0,01x=25

0×x=0

X-1/20

X-5/x²-25=0

5+x²=0

(x+3)(x-7)

Answer 3

To solve the inequality $x^2 - x + 6 > 0$ and the inequality $(x - 5)(x + 5) < 0$, we can break down the problem into two separate parts.

1. Solving $x^2 - x + 6 > 0$:

Let's start by finding the roots of the quadratic equation $x^2 - x + 6 = 0$. The roots can be found using the quadratic formula:

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

For the equation $x^2 - x + 6 = 0$, the coefficients are $a = 1$, $b = -1$, and $c = 6$. Plugging these values into the quadratic formula:

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$

Simplifying the expression under the square root:

$x = \frac{1 \pm \sqrt{-23}}{2}$

Since the expression under the square root is negative, this quadratic equation has no real roots. Therefore, the quadratic $x^2 - x + 6$ is always positive (or zero) for all real values of $x$.

2. Solving $(x - 5)(x + 5) < 0$:

We have a product of two linear factors here, $(x - 5)$ and $(x + 5)$, and we want their product to be less than zero. To do this, we need to consider the sign of each factor separately.

• When $x < -5$: Both $(x - 5)$ and $(x + 5)$ are negative, so their product $(x - 5)(x + 5)$ will be positive.

• When $-5 < x < 5$: $(x - 5)$ is negative and $(x + 5)$ is positive, so their product will be negative.

• When $x > 5$: Both $(x - 5)$ and $(x + 5)$ are positive, so their product will be positive.

So, the solution to $(x - 5)(x + 5) < 0$ is the interval $-5 < x < 5$.

To summarize:

1. The inequality $x^2 - x + 6 > 0$ is satisfied for all real values of $x$.
2. The inequality $(x - 5)(x + 5) < 0$ is satisfied when $-5 < x < 5$.

Keep in mind that these solutions are for real numbers $x$. If you're dealing with complex numbers, the analysis might be different.