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 their discriminants is

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

Answer 1

The correct option is A $a^2:p^2$

$\alpha ,\beta$ are roots of $a{x}^2+bx+c=0,a\ne 0$

$\alpha +\beta =\frac{-b}{a},\alpha \beta =\frac{c}{a}$

${△}_1=b^2-4ac=a^2(\frac{b^2}{a^2}-\frac{4c}{a})$

$=a^2({(\alpha +\beta )}^2-4\alpha \beta )$

$=a^2{(\alpha -\beta )}^2$

$\alpha +h,\beta +h$ are roots of $p{x}^2+qx+r=0,p\ne 0$

$(\alpha +h)+(\beta +h)=\frac{-q}{p},(\alpha +h)+(\beta +h)=\frac{\gamma }{p}$

${△}_2=q^2-4pr=p^2(\frac{q^2}{p^2}-\frac{4r}{p})$

$=p^2({(\gamma +h+\beta +h)}^2-4(\alpha +h)(\beta +h))$

$=p^2\left({(\alpha +h+\beta +h)}^2\right)$

${△}_2=p^2{(\alpha -\beta )}^2$

$\frac{{△}_1}{{△}_2}=\frac{b^2-4ac}{q^2-4pr}=\frac{a^2{(\alpha -\beta )}^2}{p^2{(\alpha -\beta )}^2}=\frac{a^2}{p^2}$

$\therefore {△}_1:{△}_2=a^2:p^2$