`sqrt7y^(2)-6y-13sqrt7=0`:`

Options

A. `sqrt7, 2sqrt7`
B. `3, (2)/(sqrt7)`
C. `(13)/(sqrt7), -sqrt7`
D. none of these

Answer

To solve the quadratic equation \( \sqrt{7}y^2 - 6y - 13\sqrt{7} = 0 \), we will follow these steps:

### Step 1: Identify the coefficients
The given equation is in the standard form of a quadratic equation \( ay^2 + by + c = 0 \).
- Here, \( a = \sqrt{7} \)
- \( b = -6 \)
- \( c = -13\sqrt{7} \)

### Step 2: Calculate the sum and product of the roots
For a quadratic equation \( ay^2 + by + c = 0 \):
- The sum of the roots \( (r_1 + r_2) \) is given by \( -\frac{b}{a} \)
- The product of the roots \( (r_1 \cdot r_2) \) is given by \( \frac{c}{a} \)

Calculating these:
- Sum of the roots:
\[ r_1 + r_2 = -\frac{-6}{\sqrt{7}} = \frac{6}{\sqrt{7}} \]
- Product of the roots:
\[ r_1 \cdot r_2 = \frac{-13\sqrt{7}}{\sqrt{7}} = -13 \]

### Step 3: Set up the factorization
We need to find two numbers that add up to \( \frac{6}{\sqrt{7}} \) and multiply to \( -13 \). Let’s denote these numbers as \( p \) and \( q \):
- \( p + q = \frac{6}{\sqrt{7}} \)
- \( p \cdot q = -13 \)

### Step 4: Solve for \( p \) and \( q \)
To find \( p \) and \( q \), we can set up the equations:
1. \( p + q = \frac{6}{\sqrt{7}} \)
2. \( p \cdot q = -13 \)

Let’s assume \( p = a \) and \( q = b \):
- From \( a + b = \frac{6}{\sqrt{7}} \)
- From \( ab = -13 \)

We can solve this system by substituting \( b = \frac{6}{\sqrt{7}} - a \) into the second equation:
\[ a\left(\frac{6}{\sqrt{7}} - a\right) = -13 \]
Expanding this gives:
\[ \frac{6a}{\sqrt{7}} - a^2 + 13 = 0 \]
Rearranging:
\[ a^2 - \frac{6a}{\sqrt{7}} - 13 = 0 \]

### Step 5: Use the quadratic formula
Now we can apply the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here \( A = 1 \), \( B = -\frac{6}{\sqrt{7}} \), \( C = -13 \)

Calculating the discriminant:
\[ D = \left(-\frac{6}{\sqrt{7}}\right)^2 - 4 \cdot 1 \cdot (-13) \]
\[ D = \frac{36}{7} + 52 = \frac{36 + 364}{7} = \frac{400}{7} \]

Now substituting into the quadratic formula:
\[ a = \frac{\frac{6}{\sqrt{7}} \pm \sqrt{\frac{400}{7}}}{2} \]
\[ = \frac{\frac{6}{\sqrt{7}} \pm \frac{20}{\sqrt{7}}}{2} \]
\[ = \frac{(6 \pm 20)}{2\sqrt{7}} \]

### Step 6: Find the roots
Calculating the two roots:
1. \( a_1 = \frac{26}{2\sqrt{7}} = \frac{13}{\sqrt{7}} \)
2. \( a_2 = \frac{-14}{2\sqrt{7}} = -\frac{7}{\sqrt{7}} = -\sqrt{7} \)

Thus, the roots of the equation are:
\[ y_1 = \frac{13}{\sqrt{7}} \quad \text{and} \quad y_2 = -\sqrt{7} \]

### Final Answer
The solutions to the equation \( \sqrt{7}y^2 - 6y - 13\sqrt{7} = 0 \) are:
\[ y = \frac{13}{\sqrt{7}}, \quad y = -\sqrt{7} \]

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