Question

(2×-3)^2>25 решите через Дескриминант пж​

Answer 1

Ответ:

Объяснение: x∈(3;+∞).

(2ˣ-3)²>25

2²ˣ-6*2ˣ+9-25>0

Пусть 2ˣ=t

t²-6t-16>0

t²-6t-16=0    D=100     √D=10

t₁=2ˣ=8      

t₂=2ˣ=-2    ⇒

(t-8)(t+2)>0

(2ˣ-8)(2ˣ+2)>0

2ˣ-8>0

2ˣ>2³

x>3.

Answer 2

To solve the inequality $(2x - 3)^2 > 25$ using the discriminant, we first need to rewrite the inequality in a quadratic form. Then, we can find the roots of the corresponding quadratic equation and use the discriminant to determine the solution.

Step 1: Rewrite the inequality in quadratic form:

$(2x - 3)^2 > 25$ can be simplified as follows:

$4x^2 - 12x + 9 > 25$

Step 2: Move all terms to one side to get a quadratic inequality in standard form:

$4x^2 - 12x + 9 - 25 > 0$

$4x^2 - 12x - 16 > 0$

Step 3: Now, let's find the discriminant ($D$) of the quadratic equation $ax^2 + bx + c = 0$, where $a = 4$, $b = -12$, and $c = -16$.

The discriminant is given by $D = b^2 - 4ac$.

$D = (-12)^2 - 4 * 4 * (-16) = 144 + 256 = 400$

Step 4: Analyze the discriminant and find the roots of the quadratic equation:

• If $D > 0$, there are two distinct real roots, and the parabola opens upwards (positive $a$).
• If $D = 0$, there is one real root, and the parabola is a perfect square (touches the x-axis).
• If $D < 0$, there are no real roots, and the parabola opens upwards (positive $a$).

Since $D = 400 > 0$, there are two distinct real roots.

Step 5: Find the roots using the quadratic formula:

The roots of the quadratic equation $ax^2 + bx + c = 0$ can be found using:

$x = \frac{-b \pm \sqrt{D}}{2a}$

In our case, $a = 4$ and $b = -12$:

$x = \frac{-(-12) \pm \sqrt{400}}{2 * 4}$

$x = \frac{12 \pm 20}{8}$

Now, find the two solutions:

Solution 1:

$x = \frac{12 + 20}{8} = \frac{32}{8} = 4$

Solution 2:

$x = \frac{12 - 20}{8} = \frac{-8}{8} = -1$

Step 6: Analyze the solutions:

Since we have found two distinct real roots, the parabola intersects the x-axis at $x = 4$ and $x = -1$.

Step 7: Determine the solution for the inequality:

To find the regions where the inequality $4x^2 - 12x - 16 > 0$ is satisfied, we can use a sign chart:

$$\begin{array}{c|ccccc} x & -\infty & \cdots & -1 & \cdots & 4 & \cdots & +\infty \\ \hline 4x^2-12x-16 & - & 0 & + & 0 & - \\ \end{array}$$

From the sign chart, we can see that the inequality is satisfied when $x < -1$ or $x > 4$.

So, the solution to the inequality $(2x - 3)^2 > 25$ is:

$x < -1 \text{ or } x > 4$