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Question Number 55756 by otchereabdullai@gmail.com last updated on 03/Mar/19

Find the product of all the real values of a that satisfies the equation 4∣a−4∣=∣a+4∣

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{real}\:\mathrm{values} \\ $$$$\mathrm{of}\:\:\:\mathrm{a}\:\:\mathrm{that}\:\mathrm{satisfies}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{4}\mid\mathrm{a}−\mathrm{4}\mid=\mid\mathrm{a}+\mathrm{4}\mid \\ $$

Answered by mr W last updated on 03/Mar/19

if a≤−4: −4(a−4)=−(a+4) ⇒a=((20)/3)>0 ⇒no solution if −4<a≤4 −4(a−4)=a+4 ⇒a=((12)/5)=2.4 ⇒solution if 4<a: 4(a−4)=a+4 ⇒a=((20)/3)=6.67 ⇒solution all solutions are: a=((12)/5) and ((20)/3) ⇒((12)/5)×((20)/3)=16

$${if}\:{a}\leqslant−\mathrm{4}: \\ $$$$−\mathrm{4}\left({a}−\mathrm{4}\right)=−\left({a}+\mathrm{4}\right) \\ $$$$\Rightarrow{a}=\frac{\mathrm{20}}{\mathrm{3}}>\mathrm{0}\:\Rightarrow{no}\:{solution} \\ $$$$ \\ $$$${if}\:−\mathrm{4}<{a}\leqslant\mathrm{4} \\ $$$$−\mathrm{4}\left({a}−\mathrm{4}\right)={a}+\mathrm{4} \\ $$$$\Rightarrow{a}=\frac{\mathrm{12}}{\mathrm{5}}=\mathrm{2}.\mathrm{4}\:\Rightarrow{solution} \\ $$$$ \\ $$$${if}\:\mathrm{4}<{a}: \\ $$$$\mathrm{4}\left({a}−\mathrm{4}\right)={a}+\mathrm{4} \\ $$$$\Rightarrow{a}=\frac{\mathrm{20}}{\mathrm{3}}=\mathrm{6}.\mathrm{67}\:\Rightarrow{solution} \\ $$$$ \\ $$$${all}\:{solutions}\:{are}: \\ $$$${a}=\frac{\mathrm{12}}{\mathrm{5}}\:{and}\:\frac{\mathrm{20}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{12}}{\mathrm{5}}×\frac{\mathrm{20}}{\mathrm{3}}=\mathrm{16} \\ $$

Commented by otchereabdullai@gmail.com last updated on 03/Mar/19

Thank you soo much Prof W

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{soo}\:\mathrm{much}\:\mathrm{Prof}\:\mathrm{W} \\ $$

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