Find the values of `lamda and mu` if both the roots of the equation `(3lamda+1)x^(2)=(2lamda+3mu)x-3` are infinite.

Answer

To find the values of \( \lambda \) and \( \mu \) such that both roots of the equation

\[ (3\lambda + 1)x^2 = (2\lambda + 3\mu)x - 3 \]

are infinite, we will follow these steps:

### Step 1: Rearrange the equation
First, we rearrange the equation to standard quadratic form:

\[ (3\lambda + 1)x^2 - (2\lambda + 3\mu)x + 3 = 0 \]

### Step 2: Identify coefficients
In the standard form \( ax^2 + bx + c = 0 \), we identify:
- \( a = 3\lambda + 1 \)
- \( b = -(2\lambda + 3\mu) \)
- \( c = 3 \)

### Step 3: Condition for infinite roots
For both roots to be infinite, the quadratic must have a discriminant equal to zero. The discriminant \( D \) is given by:

\[ D = b^2 - 4ac \]

Setting \( D = 0 \):

\[ (-(2\lambda + 3\mu))^2 - 4(3\lambda + 1)(3) = 0 \]

### Step 4: Simplify the discriminant
Expanding the discriminant:

\[ (2\lambda + 3\mu)^2 - 12(3\lambda + 1) = 0 \]

### Step 5: Expand and rearrange
Expanding the equation gives:

\[ 4\lambda^2 + 12\lambda\mu + 9\mu^2 - 36\lambda - 12 = 0 \]

### Step 6: Set coefficients to zero
For the quadratic to have infinite roots, we need the coefficients of the quadratic equation to be zero. Thus, we can set:

1. \( 3\lambda + 1 = 0 \)
2. \( 2\lambda + 3\mu = 0 \)

### Step 7: Solve for \( \lambda \)
From the first equation:

\[ 3\lambda + 1 = 0 \implies 3\lambda = -1 \implies \lambda = -\frac{1}{3} \]

### Step 8: Substitute \( \lambda \) into the second equation
Now substitute \( \lambda = -\frac{1}{3} \) into the second equation:

\[ 2\left(-\frac{1}{3}\right) + 3\mu = 0 \]

This simplifies to:

\[ -\frac{2}{3} + 3\mu = 0 \implies 3\mu = \frac{2}{3} \implies \mu = \frac{2}{9} \]

### Final Result
Thus, the values of \( \lambda \) and \( \mu \) are:

\[ \lambda = -\frac{1}{3}, \quad \mu = \frac{2}{9} \]