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Question Number 140908 by ajfour last updated on 14/May/21
  ab=c  let   (a−p)(b−q)=0  ⇒  c−(aq+bp)+pq=0  q=((bp−c)/(p−a))  say   4×2=8       (4−3)(2−((6−8)/(−1)))=1×0=0  And  if  q=b    p=((ab−c)/(2b))=0  And if p+q=c  (p−a)(c−p)=bp−c  p^2 −(a+c−b)p+c(1−a)=0  2p=(a+c−b)±(√((a+c−b)^2 −4c(1−a)))  ________________________
$$\:\:{ab}={c} \\ $$$${let}\:\:\:\left({a}−{p}\right)\left({b}−{q}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{c}−\left({aq}+{bp}\right)+{pq}=\mathrm{0} \\ $$$${q}=\frac{{bp}−{c}}{{p}−{a}} \\ $$$${say}\:\:\:\mathrm{4}×\mathrm{2}=\mathrm{8} \\ $$$$\:\:\:\:\:\left(\mathrm{4}−\mathrm{3}\right)\left(\mathrm{2}−\frac{\mathrm{6}−\mathrm{8}}{−\mathrm{1}}\right)=\mathrm{1}×\mathrm{0}=\mathrm{0} \\ $$$${And}\:\:{if}\:\:{q}={b} \\ $$$$\:\:{p}=\frac{{ab}−{c}}{\mathrm{2}{b}}=\mathrm{0} \\ $$$${And}\:{if}\:{p}+{q}={c} \\ $$$$\left({p}−{a}\right)\left({c}−{p}\right)={bp}−{c} \\ $$$${p}^{\mathrm{2}} −\left({a}+{c}−{b}\right){p}+{c}\left(\mathrm{1}−{a}\right)=\mathrm{0} \\ $$$$\mathrm{2}{p}=\left({a}+{c}−{b}\right)\pm\sqrt{\left({a}+{c}−{b}\right)^{\mathrm{2}} −\mathrm{4}{c}\left(\mathrm{1}−{a}\right)} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$

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