If the equations `x^(2) + 2x -3=0 and x^(2) + 3x-m=0` have a common root, then the non- zero value of m.

Answer

To solve the problem, we need to find the non-zero value of \( m \) such that the equations \( x^2 + 2x - 3 = 0 \) and \( x^2 + 3x - m = 0 \) have a common root.

### Step 1: Find the roots of the first equation
We start with the first equation:
\[ x^2 + 2x - 3 = 0 \]
To find the roots, we can factor this quadratic equation. We look for two numbers that multiply to \(-3\) (the constant term) and add to \(2\) (the coefficient of \(x\)).

The numbers \(3\) and \(-1\) satisfy these conditions:
\[ (x + 3)(x - 1) = 0 \]
Thus, the roots are:
\[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]

### Step 2: Set up the second equation with a common root
Now we have the roots \( x = -3 \) and \( x = 1 \). We will check each root in the second equation:
\[ x^2 + 3x - m = 0 \]

### Step 3: Substitute the first root \( x = -3 \)
Substituting \( x = -3 \) into the second equation:
\[ (-3)^2 + 3(-3) - m = 0 \]
Calculating this gives:
\[ 9 - 9 - m = 0 \]
\[ 0 - m = 0 \quad \Rightarrow \quad m = 0 \]
Since we are looking for a non-zero value of \( m \), we will not consider this solution.

### Step 4: Substitute the second root \( x = 1 \)
Now, substituting \( x = 1 \) into the second equation:
\[ (1)^2 + 3(1) - m = 0 \]
Calculating this gives:
\[ 1 + 3 - m = 0 \]
\[ 4 - m = 0 \quad \Rightarrow \quad m = 4 \]

### Conclusion
The non-zero value of \( m \) such that the two equations have a common root is:
\[ \boxed{4} \]