Question Number 178870 by ARUNG_Brandon_MBU last updated on 22/Oct/22
Answered by Rasheed.Sindhi last updated on 22/Oct/22
$$\mathrm{Rule}: \\ $$$$\begin{array}{|c|c|c|}{\begin{array}{|c|}{{a}}\\\hline\end{array}}&\hline{\:\:\:\:\:}&\hline{\begin{array}{|c|}{{b}}\\\hline\end{array}}\\{\:\:\:}&\hline{\begin{array}{|c|}{\frac{{ad}+{bc}}{\mathrm{2}}}\\\hline\end{array}}&\hline{\:\:}\\{\begin{array}{|c|}{{c}}\\\hline\end{array}}&\hline{\:\:\:}&\hline{\begin{array}{|c|}{{d}}\\\hline\end{array}}\\\hline\end{array}\:\:\: \\ $$$$\mathrm{Applying}\:\mathrm{rule}: \\ $$$$\begin{array}{|c|c|c|}{\begin{array}{|c|}{{x}+\mathrm{2}}\\\hline\end{array}}&\hline{\:\:\:\:\:}&\hline{\begin{array}{|c|}{{x}+\mathrm{1}}\\\hline\end{array}}\\{\:\:\:}&\hline{\begin{array}{|c|}{\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{3}\right)+\left({x}\right)\left({x}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1}}\\\hline\end{array}}&\hline{\:\:}\\{\begin{array}{|c|}{\:\:\:\:{x}\:\:\:}\\\hline\end{array}}&\hline{\:\:\:}&\hline{\begin{array}{|c|}{\:\:\:\:\mathrm{3}\:\:}\\\hline\end{array}}\\\hline\end{array} \\ $$$$\:\:\:\frac{\left({x}+\mathrm{2}\right)\left(\mathrm{3}\right)+\left({x}\right)\left({x}+\mathrm{1}\right)}{\mathrm{2}}=\mathrm{1} \\ $$$$\:\:\:\mathrm{3}{x}+\mathrm{6}+{x}^{\mathrm{2}} +{x}=\mathrm{2} \\ $$$$\:\:\:{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:{x}=−\mathrm{2} \\ $$$$\:\:\: \\ $$
Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22
Great! Thanks Sir