Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
Certainly! Let's solve the given inequality step by step:
[tex]\[ (2x + 1)^2 + 3(x + 2)^2 \geq \left(x + \frac{1}{2}\right)^2 \][/tex]
### Step 1: Expand each squared term
Expand [tex]\((2x + 1)^2\)[/tex]:
[tex]\[ (2x + 1)^2 = (2x)^2 + 2 \cdot 2x \cdot 1 + 1^2 = 4x^2 + 4x + 1 \][/tex]
Expand [tex]\(3(x + 2)^2\)[/tex]:
[tex]\[ 3(x + 2)^2 = 3[(x)^2 + 2 \cdot x \cdot 2 + 2^2] = 3(x^2 + 4x + 4) = 3x^2 + 12x + 12 \][/tex]
Expand [tex]\(\left(x + \frac{1}{2}\right)^2\)[/tex]:
[tex]\[ \left(x + \frac{1}{2}\right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2 = x^2 + x + \frac{1}{4} \][/tex]
### Step 2: Substitute the expanded terms into the inequality
Now, replace the expanded forms back into the original inequality:
[tex]\[ (4x^2 + 4x + 1) + (3x^2 + 12x + 12) \geq (x^2 + x + \frac{1}{4}) \][/tex]
Combine like terms on the left side:
[tex]\[ 4x^2 + 3x^2 + 4x + 12x + 1 + 12 \geq x^2 + x + \frac{1}{4} \][/tex]
[tex]\[ 7x^2 + 16x + 13 \geq x^2 + x + \frac{1}{4} \][/tex]
### Step 3: Bring all terms to one side of the inequality
Subtract [tex]\(x^2 + x + \frac{1}{4}\)[/tex] from both sides:
[tex]\[ 7x^2 + 16x + 13 - x^2 - x - \frac{1}{4} \geq 0 \][/tex]
Combine like terms:
[tex]\[ 6x^2 + 15x + \left(13 - \frac{1}{4}\right) \geq 0 \][/tex]
[tex]\[ 6x^2 + 15x + \frac{51}{4} \geq 0 \][/tex]
### Conclusion
The derived inequality is:
[tex]\[ 6x^2 + 15x + \frac{51}{4} \geq 0 \][/tex]
Given the solution, this inequality is satisfied for all real numbers [tex]\(x \in \mathbb{R}\)[/tex]. Therefore, the solution to the inequality is:
[tex]\[ \boxed{x \in (-\infty, \infty)} \][/tex]
[tex]\[ (2x + 1)^2 + 3(x + 2)^2 \geq \left(x + \frac{1}{2}\right)^2 \][/tex]
### Step 1: Expand each squared term
Expand [tex]\((2x + 1)^2\)[/tex]:
[tex]\[ (2x + 1)^2 = (2x)^2 + 2 \cdot 2x \cdot 1 + 1^2 = 4x^2 + 4x + 1 \][/tex]
Expand [tex]\(3(x + 2)^2\)[/tex]:
[tex]\[ 3(x + 2)^2 = 3[(x)^2 + 2 \cdot x \cdot 2 + 2^2] = 3(x^2 + 4x + 4) = 3x^2 + 12x + 12 \][/tex]
Expand [tex]\(\left(x + \frac{1}{2}\right)^2\)[/tex]:
[tex]\[ \left(x + \frac{1}{2}\right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2 = x^2 + x + \frac{1}{4} \][/tex]
### Step 2: Substitute the expanded terms into the inequality
Now, replace the expanded forms back into the original inequality:
[tex]\[ (4x^2 + 4x + 1) + (3x^2 + 12x + 12) \geq (x^2 + x + \frac{1}{4}) \][/tex]
Combine like terms on the left side:
[tex]\[ 4x^2 + 3x^2 + 4x + 12x + 1 + 12 \geq x^2 + x + \frac{1}{4} \][/tex]
[tex]\[ 7x^2 + 16x + 13 \geq x^2 + x + \frac{1}{4} \][/tex]
### Step 3: Bring all terms to one side of the inequality
Subtract [tex]\(x^2 + x + \frac{1}{4}\)[/tex] from both sides:
[tex]\[ 7x^2 + 16x + 13 - x^2 - x - \frac{1}{4} \geq 0 \][/tex]
Combine like terms:
[tex]\[ 6x^2 + 15x + \left(13 - \frac{1}{4}\right) \geq 0 \][/tex]
[tex]\[ 6x^2 + 15x + \frac{51}{4} \geq 0 \][/tex]
### Conclusion
The derived inequality is:
[tex]\[ 6x^2 + 15x + \frac{51}{4} \geq 0 \][/tex]
Given the solution, this inequality is satisfied for all real numbers [tex]\(x \in \mathbb{R}\)[/tex]. Therefore, the solution to the inequality is:
[tex]\[ \boxed{x \in (-\infty, \infty)} \][/tex]
We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.