If `sqrt(15+sqrt(60)+sqrt84+sqrt140)=sqrta+sqrtb+sqrtc`, then the value of a+b+c?

Options

A. 5
B. 20
C. 10
D. 15

Answer

To solve the equation \( \sqrt{15 + \sqrt{60} + \sqrt{84} + \sqrt{140}} = \sqrt{a} + \sqrt{b} + \sqrt{c} \), we will follow these steps:

### Step 1: Simplify the square roots on the left-hand side
We start with the expression inside the square root:
\[ \sqrt{15 + \sqrt{60} + \sqrt{84} + \sqrt{140}} \]

We can simplify the square roots:
- \( \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15} \)
- \( \sqrt{84} = \sqrt{4 \cdot 21} = 2\sqrt{21} \)
- \( \sqrt{140} = \sqrt{4 \cdot 35} = 2\sqrt{35} \)

Thus, we rewrite the expression as:
\[ \sqrt{15 + 2\sqrt{15} + 2\sqrt{21} + 2\sqrt{35}} \]

### Step 2: Group terms for simplification
Now, we can group the terms:
\[ \sqrt{15 + 2(\sqrt{15} + \sqrt{21} + \sqrt{35})} \]

### Step 3: Use the identity for squares
We recall the identity:
\[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \]
We want to express \(15\) in terms of \(a^2 + b^2 + c^2\) and \(2(ab + ac + bc)\).

### Step 4: Identify \(a\), \(b\), and \(c\)
Let’s assume:
- \( a = \sqrt{5} \)
- \( b = \sqrt{3} \)
- \( c = \sqrt{7} \)

Then:
\[ a^2 + b^2 + c^2 = 5 + 3 + 7 = 15 \]
And:
\[ ab = \sqrt{5} \cdot \sqrt{3} = \sqrt{15}, \quad ac = \sqrt{5} \cdot \sqrt{7} = \sqrt{35}, \quad bc = \sqrt{3} \cdot \sqrt{7} = \sqrt{21} \]

Thus:
\[ 2(ab + ac + bc) = 2(\sqrt{15} + \sqrt{21} + \sqrt{35}) \]

### Step 5: Combine results
Putting it all together, we have:
\[ \sqrt{15 + 2(\sqrt{15} + \sqrt{21} + \sqrt{35})} = \sqrt{5} + \sqrt{3} + \sqrt{7} \]

### Step 6: Identify \(a\), \(b\), and \(c\)
From the comparison:
- \( a = 5 \)
- \( b = 3 \)
- \( c = 7 \)

### Step 7: Calculate \(a + b + c\)
Now, we find:
\[ a + b + c = 5 + 3 + 7 = 15 \]

Thus, the final answer is:
\[ \boxed{15} \]