Question Number 55756 by otchereabdullai@gmail.com last updated on 03/Mar/19
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{values} \\ $$$$\mathrm{of}\:\:\:\mathrm{a}\:\:\mathrm{that}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{4}\mid\mathrm{a}−\mathrm{4}\mid=\mid\mathrm{a}+\mathrm{4}\mid \\ $$
Answered by mr W last updated on 03/Mar/19
$${if}\:{a}\leqslant−\mathrm{4}: \\ $$$$−\mathrm{4}\left({a}−\mathrm{4}\right)=−\left({a}+\mathrm{4}\right) \\ $$$$\Rightarrow{a}=\frac{\mathrm{20}}{\mathrm{3}}>\mathrm{0}\:\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$${if}\:−\mathrm{4}<{a}\leqslant\mathrm{4} \\ $$$$−\mathrm{4}\left({a}−\mathrm{4}\right)={a}+\mathrm{4} \\ $$$$\Rightarrow{a}=\frac{\mathrm{12}}{\mathrm{5}}=\mathrm{2}.\mathrm{4}\:\Rightarrow{solution} \\ $$$$ \\ $$$${if}\:\mathrm{4}<{a}: \\ $$$$\mathrm{4}\left({a}−\mathrm{4}\right)={a}+\mathrm{4} \\ $$$$\Rightarrow{a}=\frac{\mathrm{20}}{\mathrm{3}}=\mathrm{6}.\mathrm{67}\:\Rightarrow{solution} \\ $$$$ \\ $$$${all}\:{solutions}\:{are}: \\ $$$${a}=\frac{\mathrm{12}}{\mathrm{5}}\:{and}\:\frac{\mathrm{20}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{12}}{\mathrm{5}}×\frac{\mathrm{20}}{\mathrm{3}}=\mathrm{16} \\ $$
Commented by otchereabdullai@gmail.com last updated on 03/Mar/19
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{soo}\:\mathrm{much}\:\mathrm{Prof}\:\mathrm{W} \\ $$