Question

What will come in the place of question mark in the following question ?
\[ \sqrt{726} + \sqrt{294} + \sqrt{1176} + \sqrt{486} - \sqrt{600} = (?) \]

• A. \(31\sqrt{2}\)
• B. \(31\sqrt{6}\)
• C. \(34\sqrt{6}\)
• D. \(34\sqrt{2}\)

Hint:

When dealing with multiple large radical terms, look at the options.
The options suggest that the final answer is a multiple of either \(\sqrt{2}\) or \(\sqrt{6}\).
Test the smallest term first, such as \(294\), and divide by \(6\), which gives \(49\) (a perfect square).
This immediately alerts you that \(\sqrt{6}\) is likely the common radical factor for all terms, which dramatically speeds up factorization.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question belongs to the topic of Number Systems, specifically involving the simplification of surds and radicals.
The objective is to simplify each square root term by finding its prime factorization and isolating perfect square factors.
Once each term is expressed in its simplest radical form, we can perform addition and subtraction of like terms.

Step 2: Key Formula or Approach:
We use the fundamental property of radicals:
\[ \sqrt{a^2 \cdot b} = a\sqrt{b} \]
This helps in reducing each radical to its simplest form where \(b\) is a non-square integer.
After simplifying, like radicals (surds with the same radicand) can be added or subtracted by operating on their coefficients.

Step 3: Detailed Explanation:
Let us simplify each term individually:
First, find the prime factorization of \(726\):
\[ 726 = 6 \times 121 = 6 \times 11^2 \]
Thus, we have:
\[ \sqrt{726} = \sqrt{11^2 \times 6} = 11\sqrt{6} \]
Second, find the prime factorization of \(294\):
\[ 294 = 6 \times 49 = 6 \times 7^2 \]
Thus, we have:
\[ \sqrt{294} = \sqrt{7^2 \times 6} = 7\sqrt{6} \]
Third, find the prime factorization of \(1176\):
\[ 1176 = 6 \times 196 = 6 \times 14^2 \]
Thus, we have:
\[ \sqrt{1176} = \sqrt{14^2 \times 6} = 14\sqrt{6} \]
Fourth, find the prime factorization of \(486\):
\[ 486 = 6 \times 81 = 6 \times 9^2 \]
Thus, we have:
\[ \sqrt{486} = \sqrt{9^2 \times 6} = 9\sqrt{6} \]
Fifth, find the prime factorization of \(600\):
\[ 600 = 6 \times 100 = 6 \times 10^2 \]
Thus, we have:
\[ \sqrt{600} = \sqrt{10^2 \times 6} = 10\sqrt{6} \]
Now, substitute these simplified terms back into the original expression:
\[ 11\sqrt{6} + 7\sqrt{6} + 14\sqrt{6} + 9\sqrt{6} - 10\sqrt{6} \]
Since all the terms have the same radicand (\(6\)), we can combine their coefficients:
\[ (11 + 7 + 14 + 9 - 10)\sqrt{6} \]
\[ = (41 - 10)\sqrt{6} \]
\[ = 31\sqrt{6} \]

Step 4: Final Answer:
The simplified value that replaces the question mark is \(31\sqrt{6}\).