Question

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

• A. $x^2=0$
• B. $x^2+2bx+4=0$
• C. $x^2-2bx+4=0$
• D. $x^2-bx+1=0$

Answer 1

The correct option is C $x^2-2bx+4=0$

Since, $\alpha$ and $\beta$ are the roots of $x^2+bx+1=0$.

$\therefore \alpha +\beta =-b,\alpha \beta =1$

Now, $\left(-\alpha -\frac{1}{\beta }\right)+\left(-\beta -\frac{1}{\alpha }\right)$

$=-(\alpha +\beta )-\left(\frac{1}{\beta }+\frac{1}{\alpha }\right)=-(\alpha +\beta )-\frac{(\alpha +\beta )}{\alpha \beta }$

$=b+b=2b$

and $\left(-\alpha -\frac{1}{\beta }\right)\left(-\beta -\frac{1}{\alpha }\right)$

$=\alpha \beta +2+\frac{1}{\alpha \beta }=1+2+1=4$.

Thus, the equation whose roots are $-(\alpha +\frac{1}{\beta })$ and $-(\beta +\frac{1}{\alpha })$, is $x^2-2bx+4=0$.