If the equation k(6x^2 + 3) + rx + (2x^2 - 1) = 0 and : Quant Question Archive [LOCKED]
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# If the equation k(6x^2 + 3) + rx + (2x^2 - 1) = 0 and

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If the equation k(6x^2 + 3) + rx + (2x^2 - 1) = 0 and [#permalink]

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26 Jan 2004, 17:22
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If the equation k(6x^2 + 3) + rx + (2x^2 - 1) = 0
and 6k(2x^2 + 1) + px + (4x^2 - 2) = 0 have both the roots common, find the value of (2r - p).
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29 Jan 2004, 12:32
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Roots for first eq
(-r (+or-) sqrt( r^2 -4(3k-1)(6k+2) ) )/ 2*(6k+2)
Roots of seconf eq
(-p/2 (+or-) sqrt( p^2/4 -4(3k-1)(6k+2) ) )/ 2*(6k+2)

-r = -p/2 and r^2 = p^2/4
so r = p/2
2r-p = 0
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29 Jan 2004, 12:39
k(6x^2 + 3) + rx + (2x^2 - 1) = 0 (a)

=> 3k(2x^2+1)+rx+(2x^2 - 1) = 0 (b)

6k(2x^2 + 1) + px + (4x^2 - 2) = 0 (c)

=> 6k(2x^2+1)+px+2(2x^2 - 1) = 0 (d)

Clearly from above, 2(b) = (d), => 2r = p
Parallel equations..   [#permalink] 29 Jan 2004, 12:39
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# If the equation k(6x^2 + 3) + rx + (2x^2 - 1) = 0 and

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