Question Number 31804 by rahul 19 last updated on 15/Mar/18
$${A}\:{quadratic}\:{equation}\:{p}\left({x}\right)=\mathrm{0}\:{having} \\ $$$${coefficient}\:{of}\:{x}^{\mathrm{2}} \:{unity}\:{is}\:{such}\:{that} \\ $$$${p}\left({x}\right)=\mathrm{0}\:{and}\:{p}\left({p}\left({p}\left({x}\right)\right)\right)=\mathrm{0}\:{have}\:{a}\: \\ $$$${common}\:{root}\:{then}, \\ $$$${prove}\:{that}\::\:\:{p}\left(\mathrm{0}\right)×{p}\left(\mathrm{1}\right)=\mathrm{0}. \\ $$
Answered by MJS last updated on 15/Mar/18
![p(x)=x^2 +bx+c p(0)×p(1)=0 ⇒ p(0)=0∨p(1)=0 case 1 p(0)=0 ⇒ c=0 p(x)=(x+b)x p(p(x))=q(x) p(p(p(x)))=p(q(x))=r(x) q(x)=(p(x)+b)p(x) ⇒ ⇒ q(0)=0 r(x)=(q(x)+b)q(x) ⇒ ⇒ r(0)=0 ⇒ ⇒ 0= common root of p(x) and r(x) case 2 p(1)=0 p(x)=(x−1)(x−c) [with b=−c−1; p(x)=x^2 −(c+1)x+c] q(x)=(p(x)−1)(p(x)−c)= =p^2 (x)−(c+1)p(x)+c r(x)=[p^2 (x)−(c+1)p(x)+c−1]× ×[p^2 (x)−(c+1)p(x)]= =[...]×[p(x)−(c+1)]×p(x) ⇒ ⇒ r(1)=0 ⇒ ⇒ 1= common root of p(x) and r(x)](Q31827.png)
$${p}\left({x}\right)={x}^{\mathrm{2}} +{bx}+{c} \\ $$$${p}\left(\mathrm{0}\right)×{p}\left(\mathrm{1}\right)=\mathrm{0}\:\Rightarrow\:{p}\left(\mathrm{0}\right)=\mathrm{0}\vee{p}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{1} \\ $$$${p}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow\:{c}=\mathrm{0} \\ $$$${p}\left({x}\right)=\left({x}+{b}\right){x} \\ $$$${p}\left({p}\left({x}\right)\right)={q}\left({x}\right) \\ $$$${p}\left({p}\left({p}\left({x}\right)\right)\right)={p}\left({q}\left({x}\right)\right)={r}\left({x}\right) \\ $$$$ \\ $$$${q}\left({x}\right)=\left({p}\left({x}\right)+{b}\right){p}\left({x}\right)\:\Rightarrow\: \\ $$$$\Rightarrow\:{q}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${r}\left({x}\right)=\left({q}\left({x}\right)+{b}\right){q}\left({x}\right)\:\Rightarrow \\ $$$$\Rightarrow\:{r}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{0}=\:\mathrm{common}\:\mathrm{root}\:\mathrm{of}\:{p}\left({x}\right)\:\mathrm{and}\:{r}\left({x}\right) \\ $$$$ \\ $$$$\mathrm{case}\:\mathrm{2} \\ $$$${p}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}−{c}\right) \\ $$$$\left[\mathrm{with}\:{b}=−{c}−\mathrm{1};\:{p}\left({x}\right)={x}^{\mathrm{2}} −\left({c}+\mathrm{1}\right){x}+{c}\right] \\ $$$${q}\left({x}\right)=\left({p}\left({x}\right)−\mathrm{1}\right)\left({p}\left({x}\right)−{c}\right)= \\ $$$$={p}^{\mathrm{2}} \left({x}\right)−\left({c}+\mathrm{1}\right){p}\left({x}\right)+{c} \\ $$$${r}\left({x}\right)=\left[{p}^{\mathrm{2}} \left({x}\right)−\left({c}+\mathrm{1}\right){p}\left({x}\right)+{c}−\mathrm{1}\right]× \\ $$$$×\left[{p}^{\mathrm{2}} \left({x}\right)−\left({c}+\mathrm{1}\right){p}\left({x}\right)\right]= \\ $$$$=\left[...\right]×\left[{p}\left({x}\right)−\left({c}+\mathrm{1}\right)\right]×{p}\left({x}\right)\:\Rightarrow \\ $$$$\Rightarrow\:{r}\left(\mathrm{1}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{1}=\:\mathrm{common}\:\mathrm{root}\:\mathrm{of}\:{p}\left({x}\right)\:\mathrm{and}\:{r}\left({x}\right) \\ $$