Question

If $\alpha$ and $\beta$ are the roots of $x^2+p=0$ where p is a prime, which equation has the roots $\frac{1}{\alpha }$ and $\frac{1}{\beta }$?

• A. $\frac{1}{x^2}+\frac{1}{p}=0$
• B. $p{x}^2+1=0$
• C. $p{x}^2-1=0$
• D. $\frac{1}{x^2}-\frac{1}{p}=0$

Answer 1

The correct option is B $p{x}^2+1=0$

$x^2+p=0$

roots are $\alpha \&\beta$

sum = $\alpha +\beta =0$

product=$\alpha \beta =p$

New roots are $\frac{1}{\alpha }\&\frac{1}{\beta }$

sum = $\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=0$

product = $\frac{1}{\alpha \beta }=\frac{1}{p}$

equation

$x^2$-(sum of roots)x+product of roots = 0

$x^2-0+\frac{1}{p}=0{px}^2+1=0$