Let N be the number of quadratic equations with coefficients from {0, 1, 2, 3,..., 9} such that 0 is a solution of each equation. Then the value of N is `2^9` (b) infinite (c) 90 (d) 900

Options

A. `2^(9)`
B. infinite
C. 90
D. 900

Answer

To find the number of quadratic equations with coefficients from the set {0, 1, 2, 3, ..., 9} such that 0 is a solution of each equation, we can follow these steps:

### Step 1: Understand the form of a quadratic equation
A quadratic equation can be expressed in the standard form:
\[ ax^2 + bx + c = 0 \]
where \( a \), \( b \), and \( c \) are coefficients.

### Step 2: Apply the condition that 0 is a solution
Since 0 is a solution of the equation, we can substitute \( x = 0 \) into the equation:
\[ a(0)^2 + b(0) + c = 0 \]
This simplifies to:
\[ c = 0 \]
Thus, the quadratic equation reduces to:
\[ ax^2 + bx = 0 \]

### Step 3: Identify the constraints on coefficients
From the above simplification, we see that:
- The coefficient \( c \) must be 0.
- The coefficient \( a \) cannot be 0 (otherwise it wouldn't be a quadratic equation). Therefore, \( a \) must be chosen from the set {1, 2, 3, ..., 9}, giving us 9 possible values for \( a \).
- The coefficient \( b \) can be any value from the set {0, 1, 2, ..., 9}, giving us 10 possible values for \( b \).

### Step 4: Calculate the total number of quadratic equations
The total number of quadratic equations can be calculated by multiplying the number of choices for \( a \) and \( b \):
\[ \text{Total equations} = (\text{choices for } a) \times (\text{choices for } b) \]
\[ \text{Total equations} = 9 \times 10 = 90 \]

### Final Answer
Thus, the value of \( N \) is:
\[ N = 90 \]