Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

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

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

[Clearly, $\alpha +hand\beta +h$ 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))

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