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Question Number 188730 by universe last updated on 06/Mar/23

let p(x) = (5/3)−6x−9x^2 and Q(y) = −4y^2 −4y+((13)/2) if there exist unique pair of real number (x,y) such that p(x)×Q(y) = 20 then find the value 6x+10y = ?

$$\:\:\mathrm{let}\:{p}\left({x}\right)\:=\:\frac{\mathrm{5}}{\mathrm{3}}−\mathrm{6}{x}−\mathrm{9}{x}^{\mathrm{2}} \:\mathrm{and}\:{Q}\left({y}\right)\:=\:−\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}{y}+\frac{\mathrm{13}}{\mathrm{2}} \\ $$$$\mathrm{if}\:\mathrm{there}\:\mathrm{exist}\:\mathrm{unique}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{real}\:\mathrm{number} \\ $$$$\:\left({x},{y}\right)\:\mathrm{such}\:\mathrm{that}\:{p}\left({x}\right)×{Q}\left({y}\right)\:=\:\mathrm{20}\:\mathrm{then} \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{6}{x}+\mathrm{10}{y}\:=\:? \\ $$

Commented by mr W last updated on 06/Mar/23

p(x)×q(y)=20 has infinite solutions!

$${p}\left({x}\right)×{q}\left({y}\right)=\mathrm{20}\:{has}\:{infinite}\:{solutions}! \\ $$

Commented by Frix last updated on 07/Mar/23

@universe: You claim the anwer is 3, this means 6x+10y=3 ⇔ y=((3(1−2x))/(10)) Inserting in p(x)×Q(y)=20 gives x^4 −2x^3 −((1165x^2 )/(216))−((581x)/(324))−((1765)/(1944))=0 Which has no useable exact solutions x_1 ≈−1.35916283 x_2 ≈3.63726315 x_(3, 4) ≈−.139050158±.405363829i ...this makes no sense or does it?

$$@\mathrm{universe}: \\ $$$$\mathrm{You}\:\mathrm{claim}\:\mathrm{the}\:\mathrm{anwer}\:\mathrm{is}\:\mathrm{3},\:\mathrm{this}\:\mathrm{means} \\ $$$$\mathrm{6}{x}+\mathrm{10}{y}=\mathrm{3}\:\Leftrightarrow\:{y}=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{2}{x}\right)}{\mathrm{10}} \\ $$$$\mathrm{Inserting}\:\mathrm{in}\:{p}\left({x}\right)×{Q}\left({y}\right)=\mathrm{20}\:\mathrm{gives} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\frac{\mathrm{1165}{x}^{\mathrm{2}} }{\mathrm{216}}−\frac{\mathrm{581}{x}}{\mathrm{324}}−\frac{\mathrm{1765}}{\mathrm{1944}}=\mathrm{0} \\ $$$$\mathrm{Which}\:\mathrm{has}\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solutions} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{1}.\mathrm{35916283}\:\:\:\:\:{x}_{\mathrm{2}} \approx\mathrm{3}.\mathrm{63726315} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} \approx−.\mathrm{139050158}\pm.\mathrm{405363829i} \\ $$$$...\mathrm{this}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}\:\mathrm{or}\:\mathrm{does}\:\mathrm{it}? \\ $$

Answered by mehdee42 last updated on 06/Mar/23

−7

$$−\mathrm{7} \\ $$

Commented by universe last updated on 06/Mar/23

answer is 3

$${answer}\:{is}\:\mathrm{3} \\ $$

Commented by mehdee42 last updated on 06/Mar/23

point of max_f isA= (−(1/3),(8/3)) point of max_q is B=(−(1/2),(8/3)) therfore if f( x).q(y)=20 then there are unique x_A =−(1/3) , y=−(1/2) ⇒10x+6y=−7

$${point}\:{of}\:{max}_{{f}} \:{isA}=\:\left(−\frac{\mathrm{1}}{\mathrm{3}},\frac{\mathrm{8}}{\mathrm{3}}\right) \\ $$$${point}\:{of}\:{max}_{{q}} \:{is}\:{B}=\left(−\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{8}}{\mathrm{3}}\right) \\ $$$${therfore}\:{if}\:{f}\left(\:{x}\right).{q}\left({y}\right)=\mathrm{20}\:{then}\:{there}\:{are}\:{unique}\:{x}_{{A}} =−\frac{\mathrm{1}}{\mathrm{3}}\:,\:{y}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{10}{x}+\mathrm{6}{y}=−\mathrm{7} \\ $$

Commented by mehdee42 last updated on 06/Mar/23

?

$$? \\ $$

Answered by Frix last updated on 06/Mar/23

p(x)×Q(y)=20 ((5/3)−6x−9x^2 )(((13)/2)−4y−4y^2 )−20=0 ⇒ (1) y=−(1/2)±((3(√(10))(3x+1))/(4(√(27x^2 +18x−5)))) ⇒ x<−((3+2(√6))/9)∨((−3+2(√6))/9)<x (2) x=−(1/3)±((4(√3)(2y+1))/(9(√(8y^2 +8y−13)))) ⇒ y<−((2+(√(30)))/4)∨y((−2+(√(30)))/4)<y ⇒ Number of real solutions is infinite [Maybe you mean “rational” instead of “real”?]

$${p}\left({x}\right)×{Q}\left({y}\right)=\mathrm{20} \\ $$$$\left(\frac{\mathrm{5}}{\mathrm{3}}−\mathrm{6}{x}−\mathrm{9}{x}^{\mathrm{2}} \right)\left(\frac{\mathrm{13}}{\mathrm{2}}−\mathrm{4}{y}−\mathrm{4}{y}^{\mathrm{2}} \right)−\mathrm{20}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\:\:\:\:\:\left(\mathrm{1}\right)\:\:\:\:\:{y}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{3}\sqrt{\mathrm{10}}\left(\mathrm{3}{x}+\mathrm{1}\right)}{\mathrm{4}\sqrt{\mathrm{27}{x}^{\mathrm{2}} +\mathrm{18}{x}−\mathrm{5}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{x}<−\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{9}}\vee\frac{−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{9}}<{x} \\ $$$$\:\:\:\:\:\left(\mathrm{2}\right)\:\:\:\:\:{x}=−\frac{\mathrm{1}}{\mathrm{3}}\pm\frac{\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{2}{y}+\mathrm{1}\right)}{\mathrm{9}\sqrt{\mathrm{8}{y}^{\mathrm{2}} +\mathrm{8}{y}−\mathrm{13}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:{y}<−\frac{\mathrm{2}+\sqrt{\mathrm{30}}}{\mathrm{4}}\vee{y}\frac{−\mathrm{2}+\sqrt{\mathrm{30}}}{\mathrm{4}}<{y} \\ $$$$\Rightarrow\:\mathrm{Number}\:\mathrm{of}\:\mathrm{real}\:\mathrm{solutions}\:\mathrm{is}\:\mathrm{infinite} \\ $$$$\left[\mathrm{Maybe}\:\mathrm{you}\:\mathrm{mean}\:``\mathrm{rational}''\:\mathrm{instead}\:\mathrm{of}\:``\mathrm{real}''?\right] \\ $$

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