Question

Ig (2x²+21 x+9)-lg(2x+1)>1

Answer 1

Ответ:

решение на фотографии

Answer 2

To solve the inequality $\log(2x^2+21x+9) - \log(2x+1) > 1$, we'll use logarithm rules to simplify the expression and find the solution.

Step 1: Combine the logarithms using the quotient rule: $\log \left( \frac{2x^2+21x+9}{2x+1} \right) > 1$

Step 2: Rewrite the inequality in exponential form: $10^{\log \left( \frac{2x^2+21x+9}{2x+1} \right)} > 10^1$

Step 3: Remove the logarithm and simplify: $\frac{2x^2+21x+9}{2x+1} > 10$

Step 4: Move all terms to one side to form a quadratic inequality: $2x^2 + 21x + 9 - 10(2x+1) > 0$

Step 5: Expand and simplify: $2x^2 + 21x + 9 - 20x - 10 > 0$ $2x^2 + x - 1 > 0$

Now, we have a quadratic inequality. To solve it, we can use various methods, such as factoring, completing the square, or using the quadratic formula. However, let's use the fact that the discriminant ($b^2 - 4ac$) to determine whether the quadratic has real solutions:

For $ax^2 + bx + c > 0$, the quadratic has real solutions and is positive in the regions where $x < x_1$ or $x > x_2$ where $x_1$ and $x_2$ are the solutions of $ax^2 + bx + c = 0$, and $x_1 < x_2$.

For our quadratic $2x^2 + x - 1 > 0$, $a = 2$, $b = 1$, and $c = -1$.

The discriminant is $b^2 - 4ac = 1 - 4(2)(-1) = 1 + 8 = 9$.

Since the discriminant is positive, the quadratic has real solutions. To find those solutions, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x = \frac{-1 \pm \sqrt{9}}{2(2)}$ $x = \frac{-1 \pm 3}{4}$

So, the solutions are: $x_1 = \frac{-1 - 3}{4} = -1$ $x_2 = \frac{-1 + 3}{4} = \frac{1}{2}$

Now, we have three intervals to test: $x < -1$, $-1 < x < \frac{1}{2}$, and $x > \frac{1}{2}$.

Choose a test point from each interval and evaluate the quadratic expression $2x^2 + x - 1$:

For $x = -2$: $2(-2)^2 + (-2) - 1 = 8 - 2 - 1 = 5$

For $x = 0$: $2(0)^2 + (0) - 1 = -1$