Question

(x^2-4x-3)(2x^2-3x+1)≥0

Answer 1

Решение задания прилагаю

Answer 2

Ответ:

ответ ниже

Объяснение:

Answer 3

To solve the inequality $(x^2-4x-3)(2x^2-3x+1) \geq 0$, you'll want to find the values of $x$ for which the expression is greater than or equal to zero. To do this, follow these steps:

Step 1: Factor both quadratic expressions:

$(x^2-4x-3) = (x-3)(x+1)$

$(2x^2-3x+1) = (2x-1)(x-1)$

Step 2: Now, we can rewrite the inequality using these factored forms:

$(x-3)(x+1)(2x-1)(x-1) \geq 0$

Step 3: To determine the sign of the expression, you can use a sign chart. Find the critical values of $x$ by setting each factor equal to zero and solving for (x):

$x-3 = 0 \Rightarrow x = 3$

$x+1 = 0 \Rightarrow x = -1$

$2x-1 = 0 \Rightarrow x = 1/2$

$x-1 = 0 \Rightarrow x = 1$

Step 4: Create a sign chart with the critical values and test points within the intervals:

• $-\infty < x < -1$
• $-1 < x < 1/2$
• $1/2 < x < 1$
• $1 < x < 3$
• $3 < x < +\infty$

Step 5: Test a value within each interval to determine the sign of the expression.

For $-\infty < x < -1$, you can choose $x = -2$:

$(x-3)(x+1)(2x-1)(x-1) = (-2-3)(-2+1)(2(-2)-1)(-2-1) = (-5)(-1)(-5)(-3) > 0$

For $-1 < x < 1/2$, you can choose $x = 0$:

$(x-3)(x+1)(2x-1)(x-1) = (-3)(1)(-1)(-1) > 0$

For $1/2 < x < 1$, you can choose $x = 0.75$:

$(x-3)(x+1)(2x-1)(x-1) = (-2.25)(1.75)(0.5)(-0.25) < 0$

For $1 < x < 3$, you can choose $x = 2.5$:

$(x-3)(x+1)(2x-1)(x-1) = (-0.5)(3.5)(4)(1.5) > 0$

For $3 < x < +\infty$, you can choose $x = 4$:

$(x-3)(x+1)(2x-1)(x-1) = (1)(5)(7)(3) > 0$

Step 6: Now, combine the sign information to find where the expression is greater than or equal to zero. The expression is non-negative in the intervals $[-1, 1/2]$ and $[3, +\infty)$.

So, the solution to the inequality $(x^2-4x-3)(2x^2-3x+1) \geq 0$ is:

$-1 \leq x \leq \frac{1}{2}$ or $x \geq 3$.