2. If $$ \alpha, \beta $$ are the roots of $$ 3 x^{2}+2 x+1=0 $$, then find the equation whose roots are $$ \frac{1-\alpha}{1+\alpha} $$ and $$ \frac{1-\beta}{1+\beta} $$

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UpStudy AI · Accepted Answer

To find the equation whose roots are $$ \frac{1-\alpha}{1+\alpha} $$ and $$ \frac{1-\beta}{1+\beta} $$, where $$ \alpha $$ and $$ \beta $$ are the roots of the quadratic equation $$ 3x^2 + 2x + 1 = 0 $$, follow these steps:

### 1. Express $$ \alpha $$ in terms of $$ y $$
Let $$ y = \frac{1 - \alpha}{1 + \alpha} $$. We can solve for $$ \alpha $$ as follows:
$$
y(1 + \alpha) = 1 - \alpha \
y + y\alpha = 1 - \alpha \
\alpha(y + 1) = 1 - y \
\alpha = \frac{1 - y}{1 + y}
$$

### 2. Substitute $$ \alpha $$ into the original equation
Substitute $$ \alpha = \frac{1 - y}{1 + y} $$ into $$ 3\alpha^2 + 2\alpha + 1 = 0 $$:
$$
3\left(\frac{1 - y}{1 + y}\right)^2 + 2\left(\frac{1 - y}{1 + y}\right) + 1 = 0
$$
Multiply through by $$ (1 + y)^2 $$ to eliminate the denominator:
$$
3(1 - 2y + y^2) + 2(1 - y^2) + (1 + 2y + y^2) = 0
$$
Expand and simplify:
$$
3 - 6y + 3y^2 + 2 - 2y^2 + 1 + 2y + y^2 = 0 \
(3y^2 - 2y^2 + y^2) + (-6y + 2y) + (3 + 2 + 1) = 0 \
2y^2 - 4y + 6 = 0
$$
Divide the entire equation by 2:
$$
y^2 - 2y + 3 = 0
$$

### Verification Using Sum and Product of Roots
Alternatively, you can verify by finding the sum and product of the new roots $$ y_1 $$ and $$ y_2 $$:
- **Sum of roots**:
$$
y_1 + y_2 = 2
$$
- **Product of roots**:
$$
y_1y_2 = 3
$$
Thus, the quadratic equation is:
$$
y^2 - 2y + 3 = 0
$$

### Final Answer
$$
y^{2} - 2\,y + 3 = 0
$$
The equation with roots $$ \frac{1-\alpha}{1+\alpha} $$ and $$ \frac{1-\beta}{1+\beta} $$ is $$ y^{2} - 2\,y + 3 = 0 $$.

If are the roots of , then find the equation whose roots are and

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Step 1: Find and

Step 2: Find the new roots

Finding :

Finding :

Step 3: Form the quadratic equation

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