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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 ))

$${common}\:{root}\:{is}\:{k} \\ $$$${k}^{\mathrm{2}} +{ak}+{b}=\mathrm{0} \\ $$$${ck}^{\mathrm{2}} +\mathrm{2}{ak}−\mathrm{3}{b}=\mathrm{0} \\ $$$$\frac{{k}^{\mathrm{2}} }{\left({a}×−\mathrm{3}{b}\right)−\left(\mathrm{2}{a}×{b}\right)}=\frac{{k}}{{bc}−\left(−\mathrm{3}{b}\right)}=\frac{\mathrm{1}}{\mathrm{2}{a}−{ac}} \\ $$$$\frac{{k}^{\mathrm{2}} }{−\mathrm{5}{ab}}=\frac{{k}}{{bc}+\mathrm{3}{b}}=\frac{\mathrm{1}}{\mathrm{2}{a}−{ac}} \\ $$$${k}^{\mathrm{2}} =\frac{−\mathrm{5}{ab}}{\mathrm{2}{a}−{ac}}=\frac{\mathrm{5}{b}}{{c}−\mathrm{2}} \\ $$$${k}=\frac{{b}\left({c}+\mathrm{3}\right)}{−{a}\left({c}−\mathrm{2}\right)} \\ $$$${so} \\ $$$$\frac{\mathrm{5}{b}}{{c}−\mathrm{2}}=\frac{\left\{{b}\left({c}+\mathrm{3}\right)\right\}^{\mathrm{2}} }{\left\{−{a}\left({c}−\mathrm{2}\right)\right\}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{5}{b}}{{c}−\mathrm{2}}=\frac{{b}^{\mathrm{2}} \left({c}+\mathrm{3}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} \left({c}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$${b}=\frac{\mathrm{5}{a}^{\mathrm{2}} \left({c}−\mathrm{2}\right)}{\left({c}+\mathrm{3}\right)^{\mathrm{2}} } \\ $$

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