For which value of k will the equations `x^(2)-kx-21=0 and x^(2)-3kx+35=0` have one common root ?

Answer

To find the value of \( k \) for which the equations \( x^2 - kx - 21 = 0 \) and \( x^2 - 3kx + 35 = 0 \) have one common root, we can follow these steps:

### Step 1: Assume the common root
Let the common root be \( \alpha \). This means that \( \alpha \) satisfies both equations.

### Step 2: Substitute \( \alpha \) into the first equation
Substituting \( \alpha \) into the first equation:
\[ \alpha^2 - k\alpha - 21 = 0 \]

### Step 3: Substitute \( \alpha \) into the second equation
Substituting \( \alpha \) into the second equation:
\[ \alpha^2 - 3k\alpha + 35 = 0 \]

### Step 4: Set up the equations
Now we have two equations:
1. \( \alpha^2 - k\alpha - 21 = 0 \) (Equation 1)
2. \( \alpha^2 - 3k\alpha + 35 = 0 \) (Equation 2)

### Step 5: Subtract the first equation from the second
Subtract Equation 1 from Equation 2:
\[ (\alpha^2 - 3k\alpha + 35) - (\alpha^2 - k\alpha - 21) = 0 \]
This simplifies to:
\[ -3k\alpha + k\alpha + 35 + 21 = 0 \]
\[ -2k\alpha + 56 = 0 \]

### Step 6: Solve for \( k \)
Rearranging gives:
\[ 2k\alpha = 56 \]
\[ k\alpha = 28 \]
Thus, we can express \( \alpha \) in terms of \( k \):
\[ \alpha = \frac{28}{k} \]

### Step 7: Substitute \( \alpha \) back into Equation 1
Now substitute \( \alpha = \frac{28}{k} \) back into Equation 1:
\[ \left(\frac{28}{k}\right)^2 - k\left(\frac{28}{k}\right) - 21 = 0 \]
This simplifies to:
\[ \frac{784}{k^2} - 28 - 21 = 0 \]
\[ \frac{784}{k^2} - 49 = 0 \]

### Step 8: Solve for \( k^2 \)
Rearranging gives:
\[ \frac{784}{k^2} = 49 \]
Multiplying both sides by \( k^2 \) gives:
\[ 784 = 49k^2 \]
Dividing both sides by 49 gives:
\[ k^2 = \frac{784}{49} = 16 \]

### Step 9: Find \( k \)
Taking the square root of both sides gives:
\[ k = \pm 4 \]

### Final Answer
Thus, the values of \( k \) for which the equations have one common root are:
\[ k = 4 \quad \text{or} \quad k = -4 \]