यदि `(sqrt(28-10sqrt(3))+sqrt(7+4sqrt(3)))/(sqrt(16+6sqrt(7)))=a+bsqrt(7)` है, तब `(2a+b)=`?

Options

A. 7
B. 14
C. `15(1)/(2)`
D. `17(1)/(2)`

Answer

To solve the equation

\[ \frac{\sqrt{28 - 10\sqrt{3}} + \sqrt{7 + 4\sqrt{3}}}{\sqrt{16 + 6\sqrt{7}}} = a + b\sqrt{7} \]

we will simplify each term step by step.

### Step 1: Simplifying \(\sqrt{28 - 10\sqrt{3}}\)

We can express \(28 - 10\sqrt{3}\) in the form \((a - b\sqrt{c})^2\).

Let’s assume:

\[ \sqrt{28 - 10\sqrt{3}} = \sqrt{a} - \sqrt{b} \]

Then squaring both sides gives:

\[ 28 - 10\sqrt{3} = a + b - 2\sqrt{ab} \]

From here, we can equate the rational and irrational parts:

1. \(a + b = 28\)
2. \(-2\sqrt{ab} = -10\sqrt{3} \Rightarrow \sqrt{ab} = 5\sqrt{3} \Rightarrow ab = 75\)

Now, we need to solve the system of equations:

\[ x + y = 28 \]
\[ xy = 75 \]

Let \(x = a\) and \(y = b\). The quadratic equation is:

\[ t^2 - 28t + 75 = 0 \]

Calculating the discriminant:

\[ D = 28^2 - 4 \cdot 75 = 784 - 300 = 484 \]

Now solving for \(t\):

\[ t = \frac{28 \pm \sqrt{484}}{2} = \frac{28 \pm 22}{2} \]

This gives us:

\[ t = 25 \quad \text{or} \quad t = 3 \]

Thus, \(a = 25\) and \(b = 3\). Therefore:

\[ \sqrt{28 - 10\sqrt{3}} = 5 - \sqrt{3} \]

### Step 2: Simplifying \(\sqrt{7 + 4\sqrt{3}}\)

Similarly, we can express \(7 + 4\sqrt{3}\):

Assuming:

\[ \sqrt{7 + 4\sqrt{3}} = \sqrt{a} + \sqrt{b} \]

Squaring gives:

\[ 7 + 4\sqrt{3} = a + b + 2\sqrt{ab} \]

Equating parts:

1. \(a + b = 7\)
2. \(2\sqrt{ab} = 4\sqrt{3} \Rightarrow \sqrt{ab} = 2\sqrt{3} \Rightarrow ab = 12\)

Now we solve:

\[ x + y = 7 \]
\[ xy = 12 \]

The quadratic equation is:

\[ t^2 - 7t + 12 = 0 \]

Calculating the discriminant:

\[ D = 7^2 - 4 \cdot 12 = 49 - 48 = 1 \]

Solving gives:

\[ t = \frac{7 \pm 1}{2} = 4 \quad \text{or} \quad 3 \]

Thus, \(a = 4\) and \(b = 3\). Therefore:

\[ \sqrt{7 + 4\sqrt{3}} = 2 + \sqrt{3} \]

### Step 3: Simplifying \(\sqrt{16 + 6\sqrt{7}}\)

Assuming:

\[ \sqrt{16 + 6\sqrt{7}} = \sqrt{a} + \sqrt{b} \]

Squaring gives:

\[ 16 + 6\sqrt{7} = a + b + 2\sqrt{ab} \]

Equating parts:

1. \(a + b = 16\)
2. \(2\sqrt{ab} = 6\sqrt{7} \Rightarrow \sqrt{ab} = 3\sqrt{7} \Rightarrow ab = 63\)

Now we solve:

\[ x + y = 16 \]
\[ xy = 63 \]

The quadratic equation is:

\[ t^2 - 16t + 63 = 0 \]

Calculating the discriminant:

\[ D = 16^2 - 4 \cdot 63 = 256 - 252 = 4 \]

Solving gives:

\[ t = \frac{16 \pm 2}{2} = 9 \quad \text{or} \quad 7 \]

Thus, \(a = 9\) and \(b = 7\). Therefore:

\[ \sqrt{16 + 6\sqrt{7}} = 3 + \sqrt{7} \]

### Step 4: Putting it all together

Now substituting back into the original equation:

\[ \frac{(5 - \sqrt{3}) + (2 + \sqrt{3})}{3 + \sqrt{7}} = \frac{7}{3 + \sqrt{7}} \]

To rationalize the denominator:

\[ \frac{7(3 - \sqrt{7})}{(3 + \sqrt{7})(3 - \sqrt{7})} = \frac{7(3 - \sqrt{7})}{9 - 7} = \frac{7(3 - \sqrt{7})}{2} \]

This simplifies to:

\[ \frac{21 - 7\sqrt{7}}{2} = \frac{21}{2} - \frac{7\sqrt{7}}{2} \]

### Step 5: Identifying \(a\) and \(b\)

From the equation:

\[ a + b\sqrt{7} = \frac{21}{2} - \frac{7\sqrt{7}}{2} \]

We can identify:

\[ a = \frac{21}{2}, \quad b = -\frac{7}{2} \]

### Final Calculation

Now, we need to find \(2a + b\):

\[ 2a + b = 2 \cdot \frac{21}{2} - \frac{7}{2} = 21 - \frac{7}{2} = \frac{42}{2} - \frac{7}{2} = \frac{35}{2} \]

Thus, the final answer is:

\[ \frac{35}{2} \]