Question

If $\alpha$ and $\beta$ are the zeros of the polynomial $x^2+4x+3$, find the polynomial where zeros are $1+\frac{\beta }{\alpha }$ and $1+\frac{\alpha }{\beta }$.

Answer 1

Solution:-

$x^2+4x+3$

Since $\alpha$ and $\beta$ are the zeroes of the above polynomial.

Therefore,

$\alpha +\beta =\frac{-b}{a}=-4$

$\alpha \cdot \beta =\frac{c}{a}=3$

Now,

$(1+\frac{\alpha }{\beta })+(1+\frac{\beta }{\alpha })$

$=2+\frac{{\alpha }^2+{\beta }^2}{\alpha \cdot \beta }$

$=2+\frac{{(\alpha +\beta )}^2-2(\alpha \cdot \beta )}{\alpha \cdot \beta }$

$=2+\frac{{(-4)}^2-2\cdot 3}{3}$

$=2+\frac{16-6}{3}$

$=2+\frac{10}{3}=\frac{16}{3}$

Also,

$(1+\frac{\alpha }{\beta })(1+\frac{\beta }{\alpha })$

$=1+\frac{\alpha }{\beta }+\frac{\beta }{\alpha }+1$

$=\frac{16}{3}$

Therefore, the polynomial with zeroes $(1+\frac{\alpha }{\beta })$ and $(1+\frac{\beta }{\alpha })$ are-

$x^2-[(1+\frac{\alpha }{\beta })+(1+\frac{\beta }{\alpha })]x+(1+\frac{\alpha }{\beta })(1+\frac{\beta }{\alpha })$

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