Question Number 58387 by azizullah last updated on 22/Apr/19
Answered by mr W last updated on 22/Apr/19
$${x}=\mathrm{1}\:{is}\:{a}\:{solution}, \\ $$$$\Rightarrow{a}+\mathrm{5}=\mathrm{3} \\ $$$$\Rightarrow{a}=−\mathrm{2} \\ $$$$−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{3}=\mathrm{0} \\ $$$$−\left(\mathrm{2}{x}−\mathrm{3}\right)\left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1} \\ $$$$\Rightarrow{x}=\frac{\mathrm{3}}{\mathrm{2}}={other}\:{solution} \\ $$
Answered by Rasheed.Sindhi last updated on 22/Apr/19
$$\alpha+\beta=−\frac{\mathrm{5}}{{a}}\:,\:\:\alpha\beta=−\frac{\mathrm{3}}{{a}} \\ $$$$\alpha=\mathrm{1}\Rightarrow\beta=−\frac{\mathrm{5}}{{a}}−\mathrm{1}\:\&\:\beta=−\frac{\mathrm{3}}{{a}} \\ $$$$\:\:\:\:−\:\frac{\mathrm{5}+{a}}{{a}}=−\frac{\mathrm{3}}{{a}}\Rightarrow{a}=−\mathrm{2} \\ $$$$\:\:\:\:\beta=−\frac{\mathrm{3}}{{a}}=−\frac{\mathrm{3}}{−\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$