Question

If $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0(a\ne 0)$ and $\alpha +h,\beta +h$ are the roots of $p{x}^2+qx+r=0(p\ne 0)$, then the ratio of the squares of their discriminants is

• A. $a^2:p^2$
• B. $a:p^2$
• C. $a^2:p$
• D. $a:2p$

Answer 1

Answer: (A)

$a^2:p^2$

Given, $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$ and $\alpha +h,\beta +h$ are the roots of $p{x}^2+qx+r=0$
$\therefore \alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$
and $\alpha +h+\beta +h=-\frac{q}{p}$

$(\alpha +h)(\beta +h)=\frac{r}{p}$
Now,

$(\alpha +h)-(\beta +h)=\alpha -\beta$

$\Rightarrow {[(\alpha +h)-(\beta +h)]}^2={(\alpha -\beta )}^2$

$\Rightarrow {[(\alpha +h)+(\beta +h)]}^2-4(\alpha +h)(\beta +h)={(\alpha +\beta )}^2-4\alpha \beta$

$\Rightarrow \frac{q^2}{p^2}-\frac{4r}{p}=\frac{b^2}{a^2}-\frac{4c}{a}$

$\Rightarrow \frac{q^2-4pr}{p^2}=\frac{b^2-4ac}{a^2}$

$\therefore \frac{b^2-4ac}{q^2-4pr}=\frac{a^2}{p^2}$
Hence, the ratio of the square of their discriminants is $a^2:p^2$