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Question Number 49294 by peter frank last updated on 05/Dec/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
	$common root is k k^2 +ak+b=0 ck^2 +2ak−3b=0 (k^2 /((a×−3b)−(2a×b)))=(k/(bc−(−3b)))=(1/(2a−ac)) (k^2 /(−5ab))=(k/(bc+3b))=(1/(2a−ac)) k^2 =((−5ab)/(2a−ac))=((5b)/(c−2)) k=((b(c+3))/(−a(c−2))) so ((5b)/(c−2))=(({b(c+3)}^2 )/({−a(c−2)}^2 )) ((5b)/(c−2))=((b^2 (c+3)^2 )/(a^2 (c−2)^2 )) b=((5a^2 (c−2))/((c+3)^2 ))$
	$c o m m o n r o o t i s k$ $k^{2} + a k + b = 0$ ${c k}^{2} + 2 a k - 3 b = 0$ $\frac{k^{2}}{(a \times - 3 b) - (2 a \times b)} = \frac{k}{b c - (- 3 b)} = \frac{1}{2 a - a c}$ $\frac{k^{2}}{- 5 a b} = \frac{k}{b c + 3 b} = \frac{1}{2 a - a c}$ $k^{2} = \frac{- 5 a b}{2 a - a c} = \frac{5 b}{c - 2}$ $k = \frac{b (c + 3)}{- a (c - 2)}$ $s o$ $\frac{5 b}{c - 2} = \frac{{b (c + 3)}^{2}}{{- a (c - 2)}^{2}}$ $\frac{5 b}{c - 2} = \frac{b^{2} {(c + 3)}^{2}}{a^{2} {(c - 2)}^{2}}$ $b = \frac{5 a^{2} (c - 2)}{{(c + 3)}^{2}}$
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