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x-2-x-12-x-15-k-x-16-find-x-in-terms-of-k-gt-0-




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 by Rasheed.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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