JEE MainMathematicsToughTheory of Equations

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

Show answer and solution

k\ (6x^{2} + 3) + rx + 2x2\ –\ 1 = 0

(6k + 2)x^{2} + rx + 3k - 1 = 0

.....(i)

\Rightarrow 6k\left( 2x^{2} + 1 \right) + px + 4x^{2} - 2 = 0

\Rightarrow (12k + 4)x^{2} + px + 6k - 2 = 0

.....(ii)
Both roots common

\Rightarrow \frac{6k + 2}{12k + 4} = \frac{r}{p} = \frac{3k - 1}{6k - 2}

\Rightarrow \frac{r}{p} = \frac{1}{2}

\Rightarrow 2^{r} = p

\Rightarrow 2r - p = 0

Source: Grass-G / M-06 Mathematics Paper 14 October.docx

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