Question

If $x_1,x_2$ are the roots of $a{x}^2$ + bx + c = 0 and $x_1+d,x_2+d$ are the roots of $p{x}^2$ + qx + r = 0, d $\ne$ 0 then

• A. $\frac{a}{p+d}=\frac{b}{q+d}=\frac{c}{d+r}$
• B. $d=\frac{bp-aq}{2ap}$
• C. a + = p
• D. $b^2-4ac=q^2-4pr$

Answer 1

The correct option is B

$d=\frac{bp-aq}{2ap}$

$a{x}^2$ + bx + c = 0 has roots $x_1,x_2$
$x_1+x_2=\frac{-b}{a},x_1{x}_2=\frac{c}{a}|x_1-x_2|=\sqrt{(x_1+x_2{)}^2-4x_1{x}_2}=\sqrt{\frac{b^2}{a^2}-\frac{4c}{a}}=\frac{\sqrt{b^2-4ac}}{|a|}...\left(1\right)$
$p{x}^2$ + qx + r = 0 has roots $x_1+d_1{x}_2+d$
$\therefore (x_1+d)+(x_2+d)=-\frac{q}{p},(x_1+d)(x_2+d)=\frac{r}{p}$
$\therefore |x_1-x_2|=\frac{1}{|p|}\sqrt{q^2-4pr}...\left(2\right)$
From (1) & (2)
$\Rightarrow \frac{b^2-4ac}{a^2}=\frac{q^2-4pr}{p^2}Now2d=-\frac{-q}{p}-(x_1+x_2)2d=\frac{-aq+bp}{ap}\therefore d=\frac{-aq+bp}{2ap}$