Question

If the equations $k \left(6 x ^{2}+3\right)+ rx +2 x ^{2}-1=0$ and $6 k \left(2 x ^{2}-1\right)+ px +4 x ^{2}+2=0$ have both roots common, then the value of $(2 r-p)$ is :

• A. 0
• B. 1 / 2
• C. 1
• D. None of these

Answer 1

Answer: (A)

Given equations are
$k\left(6 x^{2}+3\right)+r x+2 x^{2}-1=0$ and
$6 k\left(2 x^{2}-1\right)+p x+4 x^{2}+2=0$
$\Rightarrow (6 k+2) x^{2}+r x+3 k-1=0\, \dots(i)$
$\Rightarrow (12 k +4) x ^{2}+ px -6 k +2=0 \,\ldots( ii )$
Let $\alpha$ and $\beta$ be the roots of both equations (i) and (ii).
$\therefore \alpha+\beta=\frac{- r }{6 k +2}$ (from (i))
and $\alpha+\beta=\frac{-p}{12 k+4}$ (from(ii))
$\therefore \quad \frac{- r }{2(1+3 k )}=\frac{- p }{4(1+3 k )} \Rightarrow \frac{- r }{2}=\frac{- p }{4}$
$\Rightarrow -2 r=-p \Rightarrow 2 r-p=0$