Question Number 210013 by otchereabdullai@gmail.com last updated on 28/Jul/24
$$\:{If}\:{f}\left({x}\right)=\:{x}^{\mathrm{2}} +{ax}+{b}.\:{if}\:{f}\left(\mathrm{1}\right)=\:\mathrm{3}\:{and}\: \\ $$$$\:{and}\:{one}\:{of}\:{the}\:{roots}\:{of}\:{the}\:{eqiation} \\ $$$$\:{f}\left({x}\right)=\:\mathrm{0}\:{doubles}\:{the}\:{other},\:{find}\:{the} \\ $$$$\:{positive}\:{values}\:{of}\:{a}\:{and}\:{b}. \\ $$
Answered by mr W last updated on 28/Jul/24
$${say}\:{the}\:{one}\:{root}\:{is}\:{p},\:{then}\:{the}\:{other} \\ $$$${root}\:{is}\:\mathrm{2}{p}. \\ $$$${p}+\mathrm{2}{p}=−{a}\:\Rightarrow{p}=−\frac{{a}}{\mathrm{3}} \\ $$$${p}×\mathrm{2}{p}={b}\:\Rightarrow\mathrm{2}{a}^{\mathrm{2}} =\mathrm{9}{b} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}^{\mathrm{2}} +{a}+{b}=\mathrm{3}\:\Rightarrow{a}+{b}=\mathrm{2} \\ $$$$\mathrm{2}{a}^{\mathrm{2}} =\mathrm{9}\left(\mathrm{2}−{a}\right) \\ $$$$\Rightarrow\mathrm{2}{a}^{\mathrm{2}} +\mathrm{9}{a}−\mathrm{18}=\mathrm{0} \\ $$$$\Rightarrow{a}=\:\frac{\mathrm{3}}{\mathrm{2}},\:−\mathrm{6}\:\left({rejected}\right) \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by otchereabdullai@gmail.com last updated on 28/Jul/24
$${thanks}\:{soo}\:{much}\:{Prof}\:{W} \\ $$