Question Number 87272 by unknown last updated on 03/Apr/20
$$\mathrm{If}\:\:\mathrm{the}\:\mathrm{equations}\:{x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0}\:\mathrm{and}\: \\ $$$${x}^{\mathrm{2}} +{bx}+{a}=\mathrm{0}\:\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root}, \\ $$$$\mathrm{then}\:\mathrm{the}\:\mathrm{numerical}\:\mathrm{value}\:\mathrm{of}\:{a}+{b}\:\mathrm{is} \\ $$
Answered by $@ty@m123 last updated on 03/Apr/20
$${Subtract}\:{the}\:{two}\:{equatios}. \\ $$$$\left({ax}+{b}\right)−\left({bx}+{a}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right){x}−\left({a}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1}\:{is}\:{the}\:{common}\:{root}. \\ $$$${Put}\:{x}=\mathrm{1}\:{in}\:{first}\left({or}\:{second}\right)\:{equation}. \\ $$$$\Rightarrow{a}+{b}=−\mathrm{1}. \\ $$
Commented by mr W last updated on 03/Apr/20
$${you}\:{meant}\:{a}+{b}=−\mathrm{1}\:{sir}. \\ $$
Commented by $@ty@m123 last updated on 03/Apr/20
$${Thanks}\:{for}\:{correction}. \\ $$