Question

If $\alpha$ and $\beta$ are the roots of the equation $3x^2+2x+1=0$, then the equation whose roots are $\alpha +{\beta }^{-1}$ and $\beta +{\alpha }^{-1}$ is

• A. $3x^2+8x+16=0$
• B. $3x^2-8x-16=0$
• C. $3x^2+8x-16=0$
• D. $x^2+8x+16=0$

Answer 1

Answer: (A)

$3x^2+8x+16=0$

Sum of the roots of the equation $3x^2+2x+1=0$

Sum of the roots $\alpha +\beta =-\frac{2}{3}$

Product of the roots $\alpha \beta =\frac{1}{3}$

Sum of the roots of unknown equation $\alpha +\frac{1}{\beta }+\beta +\frac{1}{\alpha }$

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

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

$=-\frac{2}{3}+\frac{-\frac{2}{3}}{\frac{1}{3}}$

$=-\frac{2}{3}-2\Rightarrow \frac{-8}{3}$

Product of the roots $(\alpha +\frac{1}{\beta })(\beta +\frac{1}{\alpha })$

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

Substituting $\alpha \beta =\frac{1}{3}$

$=\frac{1}{3}+2+\frac{1}{\frac{1}{3}}$

$=\frac{16}{3}$

The general form of quadratic equation is $x^2-\left(sumoftheroots\right)x+Productoftheroots=0$

Substituting the respective values, we get

$x^2-\left(\frac{-8}{3}\right)x+\frac{16}{3}=0$

Multiplying by $3$ on both sides, we get

$3x^2+8x+16=0$