Question

$\alpha$ and $\beta$ are the zeros of the polynomial $x^2+4x+3$. The polynomial whose zeros are $1+\frac{\beta }{\alpha }$ and $1+\frac{\alpha }{\beta }$ is

• A. $3x^2-16x-16$
• B. $x^2-16x-16$
• C. $x^2-16x+16$
• D. $3x^2-16x+16$

Answer 1

Answer: (D)

$3x^2-16x+16$

Since $\alpha$ and $\beta$ are the zeros of the quadratic polynomial $x^2+4x+3$

Then, $\alpha +\beta =-4,\alpha \beta =3$

Now, the of sum of the zeros of new polynomial is

$=1+\frac{\beta }{\alpha }+1+\frac{\alpha }{\beta }=\frac{\alpha \beta +{\beta }^2+\alpha \beta +{\alpha }^2}{\alpha \beta }$

$=\frac{{\alpha }^2+{\beta }^2+2\alpha \beta }{\alpha \beta }=\frac{(\alpha +\beta {)}^2}{\alpha \beta }=\frac{(-4{)}^2}{3}=\frac{16}{3}$

Also, Product of the zeros of new polynomial is

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

$=\frac{(\alpha +\beta {)}^2}{\alpha \beta }=\frac{(-4{)}^2}{3}=\frac{16}{3}$

Therefore, the required polynomial is

$k\times [x^2-(sumofthezeros)x+productofzeros]$

$\Rightarrow k\times [x^2-\frac{16}{3}x+\frac{16}{3}]$

$\Rightarrow 3\times (x^2-\frac{16}{3}x+\frac{16}{3})$ (if $k=3)$

$\Rightarrow 3x^2-16x+16$