A homogeneous polynomial of the second degree in `n` variables i.e., the expression
`phi=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)` where `a_(ij)=a_(ji)` is called a quadratic form in `n` variables `x_(1),x_(2)`….`x_(n)` if `A=[a_(ij)]_(nxn)` is
a symmetric matrix and `x=[{:(x_(1)),(x_(2)),(x_(n)):}]` then
`X^(T)AX=[X_(1)X_(2)X_(3) . . . .X_(n)][{:(a_(11),a_(12) ....a_(1n)),(a_(21),a_(22)....a_(2n)),(a_(n1),a_(n2)....a_(n n)):}][{:(x_(1)),(x_(2)),(x_(n)):}]`
`=sum_(i=1)^(n)sum_(j=1)^(n)a_(ij)x_(i)x_(j)=phi`
Matrix A is called matrix of quadratic form `phi`.
Q. If number of distinct terms in a quadratic form is 10 then number of variables in quadratic form is

Options

A. 4
B. 3
C. 5
D. can not found uniquely

Explanation

To solve the problem of finding the number of variables \( n \) in a quadratic form given that the number of distinct terms is 10, we can follow these steps:

### Step 1: Understanding the Quadratic Form
A quadratic form in \( n \) variables can be expressed as:
\[ \phi = \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} x_i x_j \]
where \( a_{ij} = a_{ji} \) (the matrix is symmetric).

### Step 2: Counting Distinct Terms
In the expression for the quadratic form, there are two types of terms:
1. **Diagonal Terms**: These are the terms where \( i = j \), which correspond to \( x_i^2 \). There are \( n \) such terms.
2. **Off-Diagonal Terms**: These are the terms where \( i \neq j \), which correspond to \( x_i x_j \). The number of distinct pairs \( (i, j) \) where \( i \neq j \) can be counted using combinations, specifically \( \binom{n}{2} \).

### Step 3: Setting Up the Equation
The total number of distinct terms in the quadratic form is given by:
\[ \text{Total Distinct Terms} = n + \binom{n}{2} \]
We know from the problem that this total is equal to 10:
\[ n + \binom{n}{2} = 10 \]

### Step 4: Expanding the Combination
The combination \( \binom{n}{2} \) can be expressed as:
\[ \binom{n}{2} = \frac{n(n-1)}{2} \]
Substituting this into our equation gives:
\[ n + \frac{n(n-1)}{2} = 10 \]

### Step 5: Clearing the Fraction
To eliminate the fraction, multiply the entire equation by 2:
\[ 2n + n(n-1) = 20 \]

### Step 6: Rearranging the Equation
This simplifies to:
\[ n^2 + n - 20 = 0 \]

### Step 7: Factoring the Quadratic
We can factor this quadratic equation:
\[ (n + 5)(n - 4) = 0 \]
Thus, the solutions for \( n \) are:
\[ n + 5 = 0 \quad \Rightarrow \quad n = -5 \quad (\text{not valid since } n \text{ must be positive}) \]
\[ n - 4 = 0 \quad \Rightarrow \quad n = 4 \]

### Step 8: Conclusion
The number of variables \( n \) in the quadratic form is:
\[ \boxed{4} \]