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Question Number 127716 by liberty last updated on 01/Jan/21

Let p and q be two positive real number such that { ((p(√p) +q(√q) = 32)),((p(√q) + q(√p) = 31)) :} find the value of ((5(p+q)?)/7)

$$\:\mathrm{Let}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{be}\:\mathrm{two}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number} \\ $$$$\mathrm{such}\:\mathrm{that}\:\begin{cases}{\mathrm{p}\sqrt{\mathrm{p}}\:+\mathrm{q}\sqrt{\mathrm{q}}\:=\:\mathrm{32}}\\{\mathrm{p}\sqrt{\mathrm{q}}\:+\:\mathrm{q}\sqrt{\mathrm{p}}\:=\:\mathrm{31}}\end{cases} \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\frac{\mathrm{5}\left(\mathrm{p}+\mathrm{q}\right)?}{\mathrm{7}} \\ $$

Answered by mindispower last updated on 01/Jan/21

((√p)+(√q))^3 =p(√p)+q(√q)+3(p(√q)+p(√q))=32+3.31=125 ⇒(√p)+(√q)=5 2nd Eq⇔(√(pq)).((√p)+(√q))=31 ⇒(√(pq))=((31)/5) p+q=((√p)+(√q))^2 −2(√(pq)) =5^2 −2.((31)/5)=((125−62)/5)=((63)/5) (5/7)(p+q)=9

$$\left(\sqrt{{p}}+\sqrt{{q}}\right)^{\mathrm{3}} ={p}\sqrt{{p}}+{q}\sqrt{{q}}+\mathrm{3}\left({p}\sqrt{{q}}+{p}\sqrt{{q}}\right)=\mathrm{32}+\mathrm{3}.\mathrm{31}=\mathrm{125} \\ $$$$\Rightarrow\sqrt{{p}}+\sqrt{{q}}=\mathrm{5} \\ $$$$\mathrm{2}{nd}\:{Eq}\Leftrightarrow\sqrt{{pq}}.\left(\sqrt{{p}}+\sqrt{{q}}\right)=\mathrm{31} \\ $$$$\Rightarrow\sqrt{{pq}}=\frac{\mathrm{31}}{\mathrm{5}} \\ $$$${p}+{q}=\left(\sqrt{{p}}+\sqrt{{q}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{{pq}} \\ $$$$=\mathrm{5}^{\mathrm{2}} −\mathrm{2}.\frac{\mathrm{31}}{\mathrm{5}}=\frac{\mathrm{125}−\mathrm{62}}{\mathrm{5}}=\frac{\mathrm{63}}{\mathrm{5}} \\ $$$$\frac{\mathrm{5}}{\mathrm{7}}\left({p}+{q}\right)=\mathrm{9} \\ $$$$ \\ $$

Answered by bemath last updated on 01/Jan/21

{ ((p(√p) + q(√q) = 32 ...(i))),((p(√q) + q(√p) = 31 ...(ii))) :} (i)+(ii) ⇒ p((√p) +(√q) )+q((√p) +(√q) )=63 ⇒(p+q)((√p) +(√q) ) = 63 ...(iii) squaring⇒(p+q)^2 ((√p) +(√q) )^2 = 63^2 ...(iv) consider eq (ii) : (√(pq)) ((√p) + (√q) ) = 31 we get ((63)/(p+q)) = ((31)/( (√(pq)))) or (√(pq)) = ((31)/(63)) (p+q)...(v) we know that ((√p) +(√q) )^2 = p+q+2(√(pq)) ((√p) +(√q) )^2 = p+q + ((62)/(63))(p+q) ((√p) +(√q) )^2 = ((125)/(63)) (p+q) ...(vi) substitute into eq (iv) gives (p+q)^2 × ((125)/(63))(p+q) = 63^2 (p+q)^3 = ((63^3 )/5^3 ) ⇒p+q = ((63)/5) therefore ((5(p+q))/7) = (5/7)×((63)/5) = 9

$$\:\begin{cases}{\mathrm{p}\sqrt{\mathrm{p}}\:+\:\mathrm{q}\sqrt{\mathrm{q}}\:=\:\mathrm{32}\:...\left(\mathrm{i}\right)}\\{\mathrm{p}\sqrt{\mathrm{q}}\:+\:\mathrm{q}\sqrt{\mathrm{p}}\:=\:\mathrm{31}\:...\left(\mathrm{ii}\right)}\end{cases} \\ $$$$\:\left(\mathrm{i}\right)+\left(\mathrm{ii}\right)\:\Rightarrow\:\mathrm{p}\left(\sqrt{\mathrm{p}}\:+\sqrt{\mathrm{q}}\:\right)+\mathrm{q}\left(\sqrt{\mathrm{p}}\:+\sqrt{\mathrm{q}}\:\right)=\mathrm{63} \\ $$$$\Rightarrow\left(\mathrm{p}+\mathrm{q}\right)\left(\sqrt{\mathrm{p}}\:+\sqrt{\mathrm{q}}\:\right)\:=\:\mathrm{63}\:...\left(\mathrm{iii}\right) \\ $$$$\mathrm{squaring}\Rightarrow\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}} \left(\sqrt{\mathrm{p}}\:+\sqrt{\mathrm{q}}\:\right)^{\mathrm{2}} \:=\:\mathrm{63}^{\mathrm{2}} \:...\left(\mathrm{iv}\right)\: \\ $$$$\mathrm{consider}\:\mathrm{eq}\:\left(\mathrm{ii}\right)\::\:\sqrt{\mathrm{pq}}\:\left(\sqrt{\mathrm{p}}\:+\:\sqrt{\mathrm{q}}\:\right)\:=\:\mathrm{31} \\ $$$$\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{63}}{\mathrm{p}+\mathrm{q}}\:=\:\frac{\mathrm{31}}{\:\sqrt{\mathrm{pq}}}\:\mathrm{or}\:\sqrt{\mathrm{pq}}\:=\:\frac{\mathrm{31}}{\mathrm{63}}\:\left(\mathrm{p}+\mathrm{q}\right)...\left(\mathrm{v}\right) \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\left(\sqrt{\mathrm{p}}\:+\sqrt{\mathrm{q}}\:\right)^{\mathrm{2}} =\:\mathrm{p}+\mathrm{q}+\mathrm{2}\sqrt{\mathrm{pq}}\: \\ $$$$\left(\sqrt{\mathrm{p}}\:+\sqrt{\mathrm{q}}\:\right)^{\mathrm{2}} \:=\:\mathrm{p}+\mathrm{q}\:+\:\frac{\mathrm{62}}{\mathrm{63}}\left(\mathrm{p}+\mathrm{q}\right)\: \\ $$$$\:\left(\sqrt{\mathrm{p}}\:+\sqrt{\mathrm{q}}\:\right)^{\mathrm{2}} \:=\:\frac{\mathrm{125}}{\mathrm{63}}\:\left(\mathrm{p}+\mathrm{q}\right)\:...\left(\mathrm{vi}\right) \\ $$$$\mathrm{substitute}\:\mathrm{into}\:\mathrm{eq}\:\left(\mathrm{iv}\right)\:\mathrm{gives} \\ $$$$\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}} \:×\:\frac{\mathrm{125}}{\mathrm{63}}\left(\mathrm{p}+\mathrm{q}\right)\:=\:\mathrm{63}^{\mathrm{2}} \\ $$$$\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{3}} \:=\:\frac{\mathrm{63}^{\mathrm{3}} }{\mathrm{5}^{\mathrm{3}} }\:\Rightarrow\mathrm{p}+\mathrm{q}\:=\:\frac{\mathrm{63}}{\mathrm{5}} \\ $$$$\mathrm{therefore}\:\:\frac{\mathrm{5}\left(\mathrm{p}+\mathrm{q}\right)}{\mathrm{7}}\:=\:\frac{\mathrm{5}}{\mathrm{7}}×\frac{\mathrm{63}}{\mathrm{5}}\:=\:\mathrm{9}\: \\ $$$$ \\ $$

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