X2+2X-|x+1|>5

Question 1

Задача Задача №2260968

А

Аноним

17 ноября 2024 08:32 · публичный вопрос

А

Аноним 17 ноября 2024 08:32

X2+2X-|x+1|>5

Answer

To solve the inequality \( x^2 + 2x - |x+1| > 5 \), we will first consider the expression inside the absolute value, which depends on the sign of \( x + 1 \).

Step 1: Analyze the absolute value

The absolute value function \( |x + 1| \) can be expressed in different ways depending on the value of \( x \):

For \( x + 1 \geq 0 \) (i.e., \( x \geq -1 \)), we have \( |x + 1| = x + 1 \).
For \( x + 1 < 0 \) (i.e., \( x < -1 \)), we have \( |x + 1| = -(x + 1) = -x - 1 \).

Step 2: Create cases based on the sign of \( x + 1 \)

We will solve the inequality in two separate cases.

Case 1: \( x \geq -1 \)

In this case, the inequality becomes: \[ x^2 + 2x - (x + 1) > 5 \] Simplifying, we obtain: \[ x^2 + 2x - x - 1 > 5 \quad \Rightarrow \quad x^2 + x - 1 > 5 \] \[ x^2 + x - 6 > 0 \] Next, we factor the quadratic: \[ (x - 2)(x + 3) > 0 \] Now, we find the critical points, which are \( x = 2 \) and \( x = -3 \). We determine the sign of the product over the intervals:

For \( x < -3 \), both factors \( (x - 2) \) and \( (x + 3) \) are negative, so the product is positive.
For \( -3 < x < 2 \), the first factor is negative, and the second is positive, so the product is negative.
For \( x > 2 \), both factors are positive, so the product is positive.

Thus, the solution for \( x \geq -1 \) is: \[ x < -3 \quad \text{or} \quad x > 2 \]

Since this case only deals with \( x \geq -1 \), the relevant solution part here is: \[ x > 2 \]

Case 2: \( x < -1 \)

For this case, the inequality becomes: \[ x^2 + 2x - (-x - 1) > 5 \] This simplifies to: \[ x^2 + 2x + x + 1 > 5 \quad \Rightarrow \quad x^2 + 3x + 1 > 5 \] \[ x^2 + 3x - 4 > 0 \] Now, we factor the quadratic: \[ (x - 1)(x + 4) > 0 \] We find the critical points which are \( x = 1 \) and \( x = -4 \). Examining the intervals:

For \( x < -4 \), both factors are negative \( \Rightarrow \) product is positive.
For \( -4 < x < 1 \), first factor is negative, second is positive \( \Rightarrow \) product is negative.
For \( x > 1 \), both factors are positive \( \Rightarrow \) product is positive.

Thus, for this case \( x < -1 \): \[ x < -4 \]

Step 3: Combine solutions

From both cases, we combine our findings:

From Case 1: \( x > 2 \)
From Case 2: \( x < -4 \)

Final Solution

The solution to the inequality \( x^2 + 2x - |x+1| > 5 \) is: \[ x < -4 \quad \text{or} \quad x > 2 \]

Question 2

X2+2X-|x+1|>5

Answer

To solve the inequality \( x^2 + 2x - |x+1| > 5 \), we will first consider the expression inside the absolute value, which depends on the sign of \( x + 1 \). ### Step 1: Analyze the absolute value The absolute value function \( |x + 1| \) can be expressed in different ways depending on the value of \( x \): - For \( x + 1 \geq 0 \) (i.e., \( x \geq -1 \)), we have \( |x + 1| = x + 1 \). - For \( x + 1 < 0 \) (i.e., \( x < -1 \)), we have \( |x + 1| = -(x + 1) = -x - 1 \). ### Step 2: Create cases based on the sign of \( x + 1 \) We will solve the inequality in two separate cases. #### Case 1: \( x \geq -1 \) In this case, the inequality becomes: \[ x^2 + 2x - (x + 1) > 5 \] Simplifying, we obtain: \[ x^2 + 2x - x - 1 > 5 \quad \Rightarrow \quad x^2 + x - 1 > 5 \] \[ x^2 + x - 6 > 0 \] Next, we factor the quadratic: \[ (x - 2)(x + 3) > 0 \] Now, we find the critical points, which are \( x = 2 \) and \( x = -3 \). We determine the sign of the product over the intervals: - For \( x < -3 \), both factors \( (x - 2) \) and \( (x + 3) \) are negative, so the product is positive. - For \( -3 < x < 2 \), the first factor is negative, and the second is positive, so the product is negative. - For \( x > 2 \), both factors are positive, so the product is positive. Thus, the solution for \( x \geq -1 \) is: \[ x < -3 \quad \text{or} \quad x > 2 \] Since this case only deals with \( x \geq -1 \), the relevant solution part here is: \[ x > 2 \] #### Case 2: \( x < -1 \) For this case, the inequality becomes: \[ x^2 + 2x - (-x - 1) > 5 \] This simplifies to: \[ x^2 + 2x + x + 1 > 5 \quad \Rightarrow \quad x^2 + 3x + 1 > 5 \] \[ x^2 + 3x - 4 > 0 \] Now, we factor the quadratic: \[ (x - 1)(x + 4) > 0 \] We find the critical points which are \( x = 1 \) and \( x = -4 \). Examining the intervals: - For \( x < -4 \), both factors are negative \( \Rightarrow \) product is positive. - For \( -4 < x < 1 \), first factor is negative, second is positive \( \Rightarrow \) product is negative. - For \( x > 1 \), both factors are positive \( \Rightarrow \) product is positive. Thus, for this case \( x < -1 \): \[ x < -4 \] ### Step 3: Combine solutions From both cases, we combine our findings: - From **Case 1**: \( x > 2 \) - From **Case 2**: \( x < -4 \) ### Final Solution The solution to the inequality \( x^2 + 2x - |x+1| > 5 \) is: \[ x < -4 \quad \text{or} \quad x > 2 \]