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Question Number 141663 by ajfour last updated on 22/May/21

  _(  −−−−−−−−−−−−−−−−−−−)     x^2 (x−12)(x−15)=k(x−16)      find x in terms of k (>0).     ^(−−−−−−−−−−−−−−−−−−−)

$$\:\underset{\:\:−−−−−−−−−−−−−−−−−−−} {\:} \\ $$ $$\:\:{x}^{\mathrm{2}} \left({x}−\mathrm{12}\right)\left({x}−\mathrm{15}\right)={k}\left({x}−\mathrm{16}\right) \\ $$ $$\:\:\:\:{find}\:{x}\:{in}\:{terms}\:{of}\:{k}\:\left(>\mathrm{0}\right). \\ $$ $$\:\:\overset{−−−−−−−−−−−−−−−−−−−} {\:} \\ $$

Commented byRasheed.Sindhi last updated on 22/May/21

((x^2 (x−12)(x−15))/(x−16))=k  k>0 ⇒x≰12 ∧ 15≰x≰16            ⇒x∈(12,15)∪(16,∞)

$$\frac{{x}^{\mathrm{2}} \left({x}−\mathrm{12}\right)\left({x}−\mathrm{15}\right)}{{x}−\mathrm{16}}={k} \\ $$ $${k}>\mathrm{0}\:\Rightarrow{x}\nleq\mathrm{12}\:\wedge\:\mathrm{15}\nleq{x}\nleq\mathrm{16} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\Rightarrow{x}\in\left(\mathrm{12},\mathrm{15}\right)\cup\left(\mathrm{16},\infty\right) \\ $$ $$ \\ $$

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