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

The correct option is A

$a^2:p^2$

Explanation for the correct option:

Step 1. Let $\alpha ,\beta$ are the roots of the equation.

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

As we know,

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

$\Rightarrow$$\alpha +h+\beta +h=\frac{–q}{p};(\alpha +h)(\beta +h)=\frac{r}{p}$

Now,

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

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

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

Step 2. Put the values of $(\alpha +h),(\beta +h),(\alpha +\beta ),\alpha \beta$ in above equation, we get

${\left(\frac{-q}{p}\right)}^2–\left(\frac{4r}{p}\right)={\left(\frac{-b}{2}\right)}^2–\left(\frac{4c}{a}\right)$

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

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

$\therefore$The ratio is ${\mathit{a}}^{\mathbf{2}}\mathbf{}\mathbf{:}\mathbf{}{\mathit{p}}^{\mathbf{2}}$

Hence, Option ‘A’ is Correct.