Question

Let $\alpha ,\beta$ be the roots of $x^2+bx+1=0$. Then find the equation whose roots are $-(\alpha +\frac{1}{\beta })$ and $-(\beta +\frac{1}{\alpha })$.

• A. $x^2-x\left(2b\right)+4=0$
• B. $x^2-2x\left(2b\right)+4=0$
• C. $x^2-x\left(4b\right)+4=0$
• D. $x^2-4x\left(4b\right)+4=0$

Answer 1

The correct option is A $x^2-x\left(2b\right)+4=0$

$-(\alpha +\beta +\frac{1}{\alpha }+\frac{1}{\beta })$
$=-(-b+\frac{\alpha +\beta }{\alpha .\beta })$
$=-(-2b)$
$=2b$
And
$(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })$
$=\alpha .\beta +1+1+\frac{1}{\alpha .\beta }$
$=1+2+1$
$=4$
Hence, the required equation will be
$x^2-2bx+4=0$