Question

If $\alpha$ and $\beta$ are the zeroes of the polynomial $x^2 + 4x + 3$, form the polynom. whose zeroes are $1 + \frac{\beta}{\alpha}$ and $1 + \frac{\alpha}{\beta}$.

Answer 1

3x^2 - 16x + 16 = 0

Answer 2

Solution Explanation

1. For the quadratic $x^2 + 4x + 3 = 0$, the sum and product of its roots are:

$$\alpha+\beta = -4,\quad \alpha\beta = 3.$$

2. The new zeros are:

$$u=1+\frac{\beta}{\alpha},\quad v=1+\frac{\alpha}{\beta}.$$

3. Their sum is:

$$u+v = 2 + \left(\frac{\beta}{\alpha}+\frac{\alpha}{\beta}\right) = 2+\frac{\alpha^2+\beta^2}{\alpha\beta}.$$

Since

$$\alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta = (-4)^2-2\cdot3 = 16-6=10,$$

we get:

$$u+v = 2+\frac{10}{3} = \frac{16}{3}.$$

4. Their product is:

$$uv=\left(1+\frac{\beta}{\alpha}\right)\left(1+\frac{\alpha}{\beta}\right)= 1+\frac{\beta}{\alpha} +\frac{\alpha}{\beta} +1 = 2+\frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{16}{3}.$$

5. Hence, the required quadratic with roots $u$ and $v$ is:

$$x^2 - \left(\frac{16}{3}\right)x + \frac{16}{3} = 0.$$

Multiplying through by 3 to eliminate the fractions gives:

$$3x^2 - 16x + 16 = 0.$$

Answer

$$\boxed{3x^2 - 16x + 16 = 0}$$