If `sqrt(10+sqrt24 +sqrt40+sqrt60)= sqrtp+sqrtq+sqrtr` then value of p + q + r is

Options

A. `sqrt10`
B. 10
C. 11
D. `sqrt10`

Answer

To solve the equation \( \sqrt{10 + \sqrt{24} + \sqrt{40} + \sqrt{60}} = \sqrt{p} + \sqrt{q} + \sqrt{r} \), we will follow these steps:

### Step 1: Simplify the square roots
First, we simplify the square roots in the expression:
- \( \sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6} \)
- \( \sqrt{40} = \sqrt{4 \cdot 10} = 2\sqrt{10} \)
- \( \sqrt{60} = \sqrt{4 \cdot 15} = 2\sqrt{15} \)

Now, substituting these back into the equation gives:
\[ \sqrt{10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}} \]

### Step 2: Set the equation
We can now rewrite the equation as:
\[ \sqrt{10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}} = \sqrt{p} + \sqrt{q} + \sqrt{r} \]

### Step 3: Square both sides
Next, we square both sides to eliminate the square root:
\[ 10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15} = (\sqrt{p} + \sqrt{q} + \sqrt{r})^2 \]

### Step 4: Expand the right side
Using the expansion formula \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc) \), we have:
\[ (\sqrt{p} + \sqrt{q} + \sqrt{r})^2 = p + q + r + 2(\sqrt{pq} + \sqrt{qr} + \sqrt{rp}) \]

### Step 5: Equate the two sides
Now we equate both sides:
\[ 10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15} = p + q + r + 2(\sqrt{pq} + \sqrt{qr} + \sqrt{rp}) \]

### Step 6: Identify terms
From this equation, we can identify:
- The constant term: \( p + q + r = 10 \)
- The terms involving square roots: \( 2(\sqrt{pq} + \sqrt{qr} + \sqrt{rp}) = 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15} \)

### Step 7: Simplify the square root terms
Dividing the square root terms by 2 gives:
\[ \sqrt{pq} + \sqrt{qr} + \sqrt{rp} = \sqrt{6} + \sqrt{10} + \sqrt{15} \]

### Step 8: Find \( p + q + r \)
From our earlier identification, we have:
\[ p + q + r = 10 \]

### Conclusion
Thus, the value of \( p + q + r \) is:
\[ \boxed{10} \]