If one root of the equation `3x^2-8x+2k+1=0` is seven times the other, find the two roots and the value of k

Answer

To solve the equation \(3x^2 - 8x + 2k + 1 = 0\) given that one root is seven times the other, we can follow these steps:

### Step 1: Define the roots
Let the first root be \( \alpha \). Then the second root will be \( 7\alpha \).

### Step 2: Use the sum of the roots
According to Vieta's formulas, the sum of the roots \( \alpha + 7\alpha \) is equal to \( -\frac{b}{a} \), where \( b = -8 \) and \( a = 3 \).

\[ \alpha + 7\alpha = 8\alpha = -\frac{-8}{3} = \frac{8}{3} \]

### Step 3: Solve for \( \alpha \)
Now, we can solve for \( \alpha \):

\[ 8\alpha = \frac{8}{3} \]

Dividing both sides by 8:

\[ \alpha = \frac{8}{3} \cdot \frac{1}{8} = \frac{1}{3} \]

### Step 4: Find the second root
Now, we can find the second root:

\[ 7\alpha = 7 \cdot \frac{1}{3} = \frac{7}{3} \]

### Step 5: Use the product of the roots
The product of the roots \( \alpha \cdot 7\alpha \) is equal to \( \frac{c}{a} \), where \( c = 2k + 1 \) and \( a = 3 \).

\[ \alpha \cdot 7\alpha = 7\alpha^2 = \frac{c}{a} \]

Substituting the values we have:

\[ 7 \cdot \left(\frac{1}{3}\right)^2 = \frac{2k + 1}{3} \]

Calculating \( 7 \cdot \frac{1}{9} \):

\[ \frac{7}{9} = \frac{2k + 1}{3} \]

### Step 6: Cross-multiply to solve for \( k \)
Cross-multiplying gives:

\[ 7 \cdot 3 = 9(2k + 1) \]

This simplifies to:

\[ 21 = 18k + 9 \]

### Step 7: Solve for \( k \)
Rearranging the equation:

\[ 18k = 21 - 9 \]
\[ 18k = 12 \]
\[ k = \frac{12}{18} = \frac{2}{3} \]

### Summary of the roots and value of \( k \)
The two roots are \( \frac{1}{3} \) and \( \frac{7}{3} \), and the value of \( k \) is \( \frac{2}{3} \).

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