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Question Number 31320 by Joel578 last updated on 06/Mar/18

Let p and q are the roots of x^2 − 2mx − 5n = 0 and m and n are the roots of x^2 − 2px − 5q = 0 If p ≠ q ≠ m ≠ n, then the value of p + q + m + n is ...

$$\mathrm{Let}\:{p}\:\mathrm{and}\:{q}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{2}{mx}\:−\:\mathrm{5}{n}\:=\:\mathrm{0} \\ $$$$\mathrm{and}\:{m}\:\mathrm{and}\:{n}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{2}{px}\:−\:\mathrm{5}{q}\:=\:\mathrm{0} \\ $$$$\mathrm{If}\:{p}\:\neq\:{q}\:\neq\:{m}\:\neq\:{n},\:\mathrm{then}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$${p}\:+\:{q}\:+\:{m}\:+\:{n}\:\mathrm{is}\:... \\ $$

Answered by MJS last updated on 06/Mar/18

(x−p)(x−q)=x^2 −2mx−5n x^2 −(p+q)x+pq=x^2 −2mx−5n p+q=2m; pq=−5n m=((p+q)/2); n=−((pq)/5) (x−m)(x−n)=x^2 −2px−5q x^2 −(m+n)x+mn=x^2 −2px−5q m+n=2p; mn=−5q ((p+q)/2)−((pq)/5)=2p⇒p=((5q)/(15+2q)) ((p+q)/2)×((pq)/5)=5q⇒q(p^2 +pq−50)=0⇒ ⇒q_1 =0; p_1 =0: not valid (p≠q) p^2 +pq−50=0 with p=((5q)/(15+2q)) q^3 −10q^2 −300q−1225=0 try all ±q with q∣1225 ⇒q_2 =−5; p_2 =−5; not valid (p≠q) q^2 −15q−225=0 q_3 =((15)/2)(1−(√5)); p_3 =(5/2)(3+(√5)) m_3 =(5/2)(3−(√5)); n_3 =((15)/2)(1+(√5)) q_4 =n_3 ; p_4 =m_3 m+n+p+q=30

$$\left({x}−{p}\right)\left({x}−{q}\right)={x}^{\mathrm{2}} −\mathrm{2}{mx}−\mathrm{5}{n} \\ $$$${x}^{\mathrm{2}} −\left({p}+{q}\right){x}+{pq}={x}^{\mathrm{2}} −\mathrm{2}{mx}−\mathrm{5}{n} \\ $$$${p}+{q}=\mathrm{2}{m};\:{pq}=−\mathrm{5}{n} \\ $$$${m}=\frac{{p}+{q}}{\mathrm{2}};\:{n}=−\frac{{pq}}{\mathrm{5}} \\ $$$$\left({x}−{m}\right)\left({x}−{n}\right)={x}^{\mathrm{2}} −\mathrm{2}{px}−\mathrm{5}{q} \\ $$$${x}^{\mathrm{2}} −\left({m}+{n}\right){x}+{mn}={x}^{\mathrm{2}} −\mathrm{2}{px}−\mathrm{5}{q} \\ $$$${m}+{n}=\mathrm{2}{p};\:{mn}=−\mathrm{5}{q} \\ $$$$\frac{{p}+{q}}{\mathrm{2}}−\frac{{pq}}{\mathrm{5}}=\mathrm{2}{p}\Rightarrow{p}=\frac{\mathrm{5}{q}}{\mathrm{15}+\mathrm{2}{q}} \\ $$$$\frac{{p}+{q}}{\mathrm{2}}×\frac{{pq}}{\mathrm{5}}=\mathrm{5}{q}\Rightarrow{q}\left({p}^{\mathrm{2}} +{pq}−\mathrm{50}\right)=\mathrm{0}\Rightarrow \\ $$$$\Rightarrow{q}_{\mathrm{1}} =\mathrm{0};\:{p}_{\mathrm{1}} =\mathrm{0}:\:\mathrm{not}\:\mathrm{valid}\:\left({p}\neq{q}\right) \\ $$$${p}^{\mathrm{2}} +{pq}−\mathrm{50}=\mathrm{0}\:\mathrm{with}\:{p}=\frac{\mathrm{5}{q}}{\mathrm{15}+\mathrm{2}{q}} \\ $$$${q}^{\mathrm{3}} −\mathrm{10}{q}^{\mathrm{2}} −\mathrm{300}{q}−\mathrm{1225}=\mathrm{0} \\ $$$$\mathrm{try}\:\mathrm{all}\:\pm{q}\:\mathrm{with}\:{q}\mid\mathrm{1225} \\ $$$$\Rightarrow{q}_{\mathrm{2}} =−\mathrm{5};\:{p}_{\mathrm{2}} =−\mathrm{5};\:\mathrm{not}\:\mathrm{valid}\:\left({p}\neq{q}\right) \\ $$$${q}^{\mathrm{2}} −\mathrm{15}{q}−\mathrm{225}=\mathrm{0} \\ $$$${q}_{\mathrm{3}} =\frac{\mathrm{15}}{\mathrm{2}}\left(\mathrm{1}−\sqrt{\mathrm{5}}\right);\:{p}_{\mathrm{3}} =\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{3}+\sqrt{\mathrm{5}}\right) \\ $$$${m}_{\mathrm{3}} =\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{3}−\sqrt{\mathrm{5}}\right);\:{n}_{\mathrm{3}} =\frac{\mathrm{15}}{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right) \\ $$$${q}_{\mathrm{4}} ={n}_{\mathrm{3}} ;\:{p}_{\mathrm{4}} ={m}_{\mathrm{3}} \\ $$$${m}+{n}+{p}+{q}=\mathrm{30} \\ $$

Commented by Joel578 last updated on 06/Mar/18

thank you very much

$${thank}\:{you}\:{very}\:{much} \\ $$

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