Question Number 25187 by lucky singh last updated on 05/Dec/17
$${if}\:{in}\:{a}\:{quadratic}\:{eqution}\:{x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0} \\ $$$${and}\:{x}^{\mathrm{2}} +{bx}+{a}=\mathrm{0}\:{have}\:{a}\:{common}\:{root} \\ $$$${then}\:{the}\:{numerical}\:{value}\:{of}\:\left({a}+{b}\right)\:{is} \\ $$
Commented by prakash jain last updated on 05/Dec/17
$$\mathrm{let}\:\alpha\:\mathrm{be}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root} \\ $$$$\alpha^{\mathrm{2}} +{a}\alpha+{b}=\mathrm{0}\:\:\:\:\left({i}\right) \\ $$$$\alpha^{\mathrm{2}} +{b}\alpha+{a}=\mathrm{0}\:\:\:\:\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right) \\ $$$$\left({a}−{b}\right)\alpha+{b}−{a}=\mathrm{0} \\ $$$$\Rightarrow\alpha=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{put}\:{x}=\mathrm{1}\:\mathrm{in}\:{x}^{\mathrm{2}} +{ax}+{b}=\mathrm{0} \\ $$$${a}+{b}=−\mathrm{1} \\ $$