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Question Number 112133 by shwebotetun last updated on 06/Sep/20

Answered by som(math1967) last updated on 06/Sep/20

$${(-2\mathrm{b})}^2+20\mathrm{b}+14={(-2\mathrm{c})}^2+20\mathrm{c}+14$$$${4\mathrm{b}}^2-{4\mathrm{c}}^2+20(\mathrm{b}-\mathrm{c})=0$$$$4\{(\mathrm{b}-\mathrm{c})(\mathrm{b}+\mathrm{c})+5(\mathrm{b}-\mathrm{c})\}=0$$$$(\mathrm{b}-\mathrm{c})(\mathrm{b}+\mathrm{c}+5)=0$$$$\therefore \mathrm{b}+\mathrm{c}+5=0[\because \mathrm{b}\ne \mathrm{c}\therefore (\mathrm{b}-\mathrm{c})\ne 0]$$

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