Question

If the roots of $a{x}^2+bx+c=0,\ne 0,$ are p,q ($p\ne q$), then the roots of $c{x}^2-bx+a=0$ are.

• A. $p+q,p-q$
• B. $\frac{1}{p+q},\frac{1}{p-q}$
• C. $\frac{1}{p},\frac{1}{q}$
• D. $-\frac{1}{p},-\frac{1}{q}$

Answer 1

Answer: (D)

$-\frac{1}{p},-\frac{1}{q}$

Consider
$a{x}^2+bx+c=0$
The roots are
$p,q$
Hence
$p+q=\frac{-b}{a}$ and $pq=\frac{c}{a}$
Now consider
$c{x}^2-bx+a=0$
$\frac{b}{c}=\frac{\frac{b}{a}}{\frac{c}{a}}$
$=\frac{-(p+q)}{pq}$
$=-(\frac{p}{pq}+\frac{q}{pq})$
$=-(\frac{1}{q}+\frac{1}{p})$
Hence sum of roots of the new equation is
$=-\frac{1}{q}-\frac{1}{p}$
Therefore, the roots are
$\frac{-1}{p},\frac{-1}{q}$.