Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Discover a wealth of knowledge from experts across different disciplines on our comprehensive Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To find the square of the given expression [tex]\((4 x \sqrt{5 x^2} + 2 x^2 \sqrt{6})^2\)[/tex], let's break it down and simplify step-by-step.
1. Understand the individual terms inside the expression:
[tex]\[ 4 x \sqrt{5 x^2} + 2 x^2 \sqrt{6} \][/tex]
2. Simplify [tex]\( \sqrt{5 x^2} \)[/tex]:
[tex]\[ \sqrt{5 x^2} = x \sqrt{5} \quad \text{(since \(x \geq 0\))} \][/tex]
3. Substitute back into the expression:
[tex]\[ 4 x \sqrt{5 x^2} = 4 x \cdot x \sqrt{5} = 4 x^2 \sqrt{5} \][/tex]
So, the expression becomes:
[tex]\[ 4 x^2 \sqrt{5} + 2 x^2 \sqrt{6} \][/tex]
4. Factor out the common term [tex]\(2 x^2\)[/tex]:
[tex]\[ 2 x^2 (2 \sqrt{5} + \sqrt{6}) \][/tex]
5. Square the entire expression:
[tex]\[ (2 x^2 (2 \sqrt{5} + \sqrt{6}))^2 \][/tex]
6. Apply the square to each part of the product:
[tex]\[ (2 x^2)^2 \cdot (2 \sqrt{5} + \sqrt{6})^2 \][/tex]
[tex]\[ 4 x^4 \cdot (2 \sqrt{5} + \sqrt{6})^2 \][/tex]
7. Expand the binomial [tex]\((2 \sqrt{5} + \sqrt{6})^2\)[/tex]:
[tex]\[ (2 \sqrt{5} + \sqrt{6})^2 = (2 \sqrt{5})^2 + 2 \cdot (2 \sqrt{5}) \cdot (\sqrt{6}) + (\sqrt{6})^2 \][/tex]
[tex]\[ = 4 \cdot 5 + 2 \cdot 2 \sqrt{5} \cdot \sqrt{6} + 6 \][/tex]
[tex]\[ = 20 + 4 \sqrt{30} + 6 \][/tex]
[tex]\[ = 26 + 4 \sqrt{30} \][/tex]
8. Combine these results:
[tex]\[ 4 x^4 \cdot (26 + 4 \sqrt{30}) \][/tex]
9. Distribute [tex]\(4 x^4\)[/tex] through the terms inside the parentheses:
[tex]\[ 4 x^4 \cdot 26 + 4 x^4 \cdot 4 \sqrt{30} \][/tex]
[tex]\[ 104 x^4 + 16 x^4 \sqrt{30} \][/tex]
Therefore, the simplified form of [tex]\( (4 x \sqrt{5 x^2} + 2 x^2 \sqrt{6})^2 \)[/tex] is:
[tex]\[ 104 x^4 + 16 x^4 \sqrt{30} \][/tex]
Thus, the correct option is:
[tex]\[ 104 x^4 + 16 x^4 \sqrt{30} \][/tex]
So, the answer is the fourth option: [tex]\( \boxed{104 x^4 + 16 x^4 \sqrt{30}} \)[/tex].
1. Understand the individual terms inside the expression:
[tex]\[ 4 x \sqrt{5 x^2} + 2 x^2 \sqrt{6} \][/tex]
2. Simplify [tex]\( \sqrt{5 x^2} \)[/tex]:
[tex]\[ \sqrt{5 x^2} = x \sqrt{5} \quad \text{(since \(x \geq 0\))} \][/tex]
3. Substitute back into the expression:
[tex]\[ 4 x \sqrt{5 x^2} = 4 x \cdot x \sqrt{5} = 4 x^2 \sqrt{5} \][/tex]
So, the expression becomes:
[tex]\[ 4 x^2 \sqrt{5} + 2 x^2 \sqrt{6} \][/tex]
4. Factor out the common term [tex]\(2 x^2\)[/tex]:
[tex]\[ 2 x^2 (2 \sqrt{5} + \sqrt{6}) \][/tex]
5. Square the entire expression:
[tex]\[ (2 x^2 (2 \sqrt{5} + \sqrt{6}))^2 \][/tex]
6. Apply the square to each part of the product:
[tex]\[ (2 x^2)^2 \cdot (2 \sqrt{5} + \sqrt{6})^2 \][/tex]
[tex]\[ 4 x^4 \cdot (2 \sqrt{5} + \sqrt{6})^2 \][/tex]
7. Expand the binomial [tex]\((2 \sqrt{5} + \sqrt{6})^2\)[/tex]:
[tex]\[ (2 \sqrt{5} + \sqrt{6})^2 = (2 \sqrt{5})^2 + 2 \cdot (2 \sqrt{5}) \cdot (\sqrt{6}) + (\sqrt{6})^2 \][/tex]
[tex]\[ = 4 \cdot 5 + 2 \cdot 2 \sqrt{5} \cdot \sqrt{6} + 6 \][/tex]
[tex]\[ = 20 + 4 \sqrt{30} + 6 \][/tex]
[tex]\[ = 26 + 4 \sqrt{30} \][/tex]
8. Combine these results:
[tex]\[ 4 x^4 \cdot (26 + 4 \sqrt{30}) \][/tex]
9. Distribute [tex]\(4 x^4\)[/tex] through the terms inside the parentheses:
[tex]\[ 4 x^4 \cdot 26 + 4 x^4 \cdot 4 \sqrt{30} \][/tex]
[tex]\[ 104 x^4 + 16 x^4 \sqrt{30} \][/tex]
Therefore, the simplified form of [tex]\( (4 x \sqrt{5 x^2} + 2 x^2 \sqrt{6})^2 \)[/tex] is:
[tex]\[ 104 x^4 + 16 x^4 \sqrt{30} \][/tex]
Thus, the correct option is:
[tex]\[ 104 x^4 + 16 x^4 \sqrt{30} \][/tex]
So, the answer is the fourth option: [tex]\( \boxed{104 x^4 + 16 x^4 \sqrt{30}} \)[/tex].
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.