Question

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

• A. $\frac{1}{2}(\frac{q}{p}-\frac{b}{a})$
• B. $h=\frac{1}{2}(\frac{b}{a}-\frac{q}{p})$
• C. $h=\frac{1}{2}(\frac{b}{a}+\frac{q}{p})$
• D. $\frac{a}{p}=\frac{b}{q}=\frac{c}{r}$

Answer 1

The correct option is B

$h=\frac{1}{2}(\frac{b}{a}-\frac{q}{p})$

The equation whose roots are $\alpha +h$ and $\beta +h$ is obtained by replacing $x$ by $x-h$.

i.e. ${a(x-h)}^2+b(x-h)+c=0$

[Clearly, $\alpha +h$ and $\beta +h$ are the roots of this equation]

$\Rightarrow a(x^2-2hx+h^2)+bx-bh+c=0$

$\Rightarrow a{x}^2+(b-2ah)x+a{h}^2-bh+c=0$ ---------(i)

This is same as $p{x}^2+qx+r=0$ ------(ii)

Comparing both the equations we get (if two equations have the same roots or represent the same equation, then

the ratio of the corresponding coefficients is same)

$\frac{a}{p}=\frac{b-2ha}{q}=\frac{a{h}^2-bh+c}{r}$ ---------------------------(iii)

$\Rightarrow$ A is not correct.

$\Rightarrow aq=pb-2hap$ (by considering first two terms in (iii))

$\Rightarrow 2ahp=bp-aq$
$h=\frac{1}{2}\left(\frac{bp-aq}{ap}\right)=\frac{1}{2}(\frac{b}{a}-\frac{q}{p})$