Let `alpha,beta`, are two real solution of equation `(log_(10)x)^2 + log_(10)x^2 = (log_(10))^2 -1,` then ` sqrt1/(alpha beta)`equal to `(i) 20 (ii) 3 (iii) 10 (iv) 1`

Options

A. `20`
B. `3`
C. `10`
D. `1`

Answer

To solve the equation \((\log_{10} x)^2 + \log_{10} x^2 = \log_{10}^2 - 1\), we will follow these steps:

### Step 1: Rewrite the equation
We know that \(\log_{10} x^2 = 2 \log_{10} x\). Therefore, we can rewrite the equation as:
\[ (\log_{10} x)^2 + 2 \log_{10} x = \log_{10}^2 - 1 \]
This simplifies to:
\[ (\log_{10} x)^2 + 2 \log_{10} x + 1 - \log_{10}^2 = 0 \]

### Step 2: Substitute variables
Let \(t = \log_{10} x\). The equation now becomes:
\[ t^2 + 2t + 1 - \log_{10}^2 = 0 \]
Rearranging gives:
\[ t^2 + 2t + 1 = \log_{10}^2 \]

### Step 3: Factor the left-hand side
Notice that \(t^2 + 2t + 1\) can be factored as:
\[ (t + 1)^2 = \log_{10}^2 \]

### Step 4: Apply the difference of squares
Using the identity \(a^2 - b^2 = (a - b)(a + b)\), we can write:
\[ (t + 1 - \log_{10})(t + 1 + \log_{10}) = 0 \]

### Step 5: Solve for \(t\)
This gives us two equations:
1. \(t + 1 - \log_{10} = 0\) which simplifies to \(t = \log_{10} 10 - 1 = -1 + 1 = 0\)
2. \(t + 1 + \log_{10} = 0\) which simplifies to \(t = -1 - \log_{10} = -1 - \log_{10} 10 = -1 - 1 = -2\)

### Step 6: Convert back to \(x\)
Now we convert back to \(x\) using \(t = \log_{10} x\):
1. For \(t = 0\), we have \(\log_{10} x = 0\) which gives \(x = 10^0 = 1\).
2. For \(t = -2\), we have \(\log_{10} x = -2\) which gives \(x = 10^{-2} = \frac{1}{100}\).

Thus, the solutions are:
\(\alpha = 1\) and \(\beta = \frac{1}{100}\).

### Step 7: Calculate \(\sqrt{\frac{1}{\alpha \beta}}\)
Now we need to calculate:
\[ \sqrt{\frac{1}{\alpha \beta}} = \sqrt{\frac{1}{1 \cdot \frac{1}{100}}} = \sqrt{100} = 10 \]

### Final Answer
Thus, \(\sqrt{\frac{1}{\alpha \beta}} = 10\).