For what value of k will be 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: Define the common root
Let the common root be \( \alpha \). Since \( \alpha \) is a root of both equations, it must satisfy both equations:
1. \( \alpha^2 - k\alpha - 21 = 0 \) (Equation 1)
2. \( \alpha^2 - 3k\alpha + 35 = 0 \) (Equation 2)

### Step 2: Set the equations equal
From both equations, we can express them in terms of \( \alpha \):
- From Equation 1:
\[ \alpha^2 = k\alpha + 21 \]
- From Equation 2:
\[ \alpha^2 = 3k\alpha - 35 \]

### Step 3: Equate the two expressions for \( \alpha^2 \)
Since both expressions equal \( \alpha^2 \), we can set them equal to each other:
\[ k\alpha + 21 = 3k\alpha - 35 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ 21 + 35 = 3k\alpha - k\alpha \]
\[ 56 = 2k\alpha \]

### Step 5: Solve for \( \alpha \)
From the equation \( 56 = 2k\alpha \), we can express \( \alpha \) in terms of \( k \):
\[ \alpha = \frac{56}{2k} = \frac{28}{k} \]

### Step 6: Substitute \( \alpha \) back into one of the original equations
Substituting \( \alpha = \frac{28}{k} \) 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 7: Solve for \( k^2 \)
Multiplying through by \( k^2 \) to eliminate the fraction:
\[ 784 - 49k^2 = 0 \]
\[ 49k^2 = 784 \]
\[ k^2 = \frac{784}{49} = 16 \]

### Step 8: 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:
\[ k = 4 \quad \text{or} \quad k = -4 \]

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