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Question Number 83791 by jagoll last updated on 06/Mar/20
Let x, y are two different real numbers satisfy the equation (√(y+4)) = x−4 and (√(x+4)) = y−4. The value of x^3 +y^3 mod(x^3 y^3 ) is
$$\mathrm{Let}\:\mathrm{x},\:\mathrm{y}\:\mathrm{are}\:\mathrm{two}\:\mathrm{different}\:\mathrm{real} \\ $$$$\mathrm{numbers}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\sqrt{\mathrm{y}+\mathrm{4}}\:=\:\mathrm{x}−\mathrm{4}\:\mathrm{and}\:\sqrt{\mathrm{x}+\mathrm{4}}\:=\:\mathrm{y}−\mathrm{4}. \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} \:\mathrm{mod}\left(\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{3}} \right)\:\mathrm{is} \\ $$
Commented by mr W last updated on 06/Mar/20
there are no real numbers x,y which satisfy the equations and x≠y. the only real solution of the equations is x=y=((9+(√(33)))/2).
$${there}\:{are}\:{no}\:{real}\:{numbers}\:{x},{y}\:{which} \\ $$$${satisfy}\:{the}\:{equations}\:{and}\:{x}\neq{y}. \\ $$$${the}\:{only}\:{real}\:{solution}\:{of}\:{the}\:{equations} \\ $$$${is}\:{x}={y}=\frac{\mathrm{9}+\sqrt{\mathrm{33}}}{\mathrm{2}}. \\ $$
Commented by jagoll last updated on 06/Mar/20
is this problem wrong?
$$\mathrm{is}\:\mathrm{this}\:\mathrm{problem}\:\mathrm{wrong}? \\ $$
Answered by MJS last updated on 06/Mar/20
due to symmetry I believe x=y (√(x+4))=x−4 x≥4 x+4=(x−4)^2 x^2 −9x+12=0∧x≥4 ⇒ x=y=((9+(√(33)))/2) I don′t think there′s another real solution
$$\mathrm{due}\:\mathrm{to}\:\mathrm{symmetry}\:\mathrm{I}\:\mathrm{believe}\:{x}={y} \\ $$$$\sqrt{{x}+\mathrm{4}}={x}−\mathrm{4} \\ $$$${x}\geqslant\mathrm{4} \\ $$$${x}+\mathrm{4}=\left({x}−\mathrm{4}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{12}=\mathrm{0}\wedge{x}\geqslant\mathrm{4}\:\Rightarrow \\ $$$${x}={y}=\frac{\mathrm{9}+\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{another}\:\mathrm{real}\:\mathrm{solution} \\ $$
Commented by MJS last updated on 06/Mar/20
...there′s no other solution in C
$$…\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{other}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{C} \\ $$
Commented by jagoll last updated on 06/Mar/20
answer choices (a) 3 (b) 13 (c)103 (d) 113 (e) 131 sir
$$\mathrm{answer}\:\mathrm{choices}\:\left(\mathrm{a}\right)\:\mathrm{3}\:\:\left(\mathrm{b}\right)\:\mathrm{13}\:\left(\mathrm{c}\right)\mathrm{103} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{113}\:\left(\mathrm{e}\right)\:\mathrm{131}\:\mathrm{sir}\: \\ $$
Answered by MJS last updated on 06/Mar/20
if x∈R∧y∈R∧x≠y ⇒ y=x+t∧t∈R (1) (√(x+t+4))=x−4 (2) (√(x+4))=x+t−4 x+t+4≥0∧x−4≥0∧x+4≥0∧x+t−4≥0 ⇒ ⇒ x≥4∧t≥4−x (2)−(1) (√(x+t+4))−(√(x+4))=t we have to square ⇒ beware of false solutions! ⇒ t=0∨t=1+2(√(x+4)) if t=0 ⇒ y=x if t=1+2(√(x+4)) ⇒ y=x+1+2(√(x+4)) (1) (√(x+5+2(√(x+4))))=x−4 ⇒x=((11+(√(37)))/2) (2) (√(x+4))=x−3+2(√(x+4)) ⇒ (√(x+4))=3−x ⇒ ⇒ x=((7−(√(29)))/2) ⇒ wrong ⇒ t=0 ⇒ y=x
$$\mathrm{if}\:{x}\in\mathbb{R}\wedge{y}\in\mathbb{R}\wedge{x}\neq{y}\:\Rightarrow\:{y}={x}+{t}\wedge{t}\in\mathbb{R} \\ $$$$\left(\mathrm{1}\right)\:\sqrt{{x}+{t}+\mathrm{4}}={x}−\mathrm{4} \\ $$$$\left(\mathrm{2}\right)\:\sqrt{{x}+\mathrm{4}}={x}+{t}−\mathrm{4} \\ $$$${x}+{t}+\mathrm{4}\geqslant\mathrm{0}\wedge{x}−\mathrm{4}\geqslant\mathrm{0}\wedge{x}+\mathrm{4}\geqslant\mathrm{0}\wedge{x}+{t}−\mathrm{4}\geqslant\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{x}\geqslant\mathrm{4}\wedge{t}\geqslant\mathrm{4}−{x} \\ $$$$\left(\mathrm{2}\right)−\left(\mathrm{1}\right)\:\sqrt{{x}+{t}+\mathrm{4}}−\sqrt{{x}+\mathrm{4}}={t} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{square}\:\Rightarrow\:\mathrm{beware}\:\mathrm{of}\:\mathrm{false}\:\mathrm{solutions}! \\ $$$$\Rightarrow\:{t}=\mathrm{0}\vee{t}=\mathrm{1}+\mathrm{2}\sqrt{{x}+\mathrm{4}} \\ $$$$\mathrm{if}\:{t}=\mathrm{0}\:\Rightarrow\:{y}={x} \\ $$$$ \\ $$$$\mathrm{if}\:{t}=\mathrm{1}+\mathrm{2}\sqrt{{x}+\mathrm{4}}\:\Rightarrow\:{y}={x}+\mathrm{1}+\mathrm{2}\sqrt{{x}+\mathrm{4}} \\ $$$$\left(\mathrm{1}\right)\:\sqrt{{x}+\mathrm{5}+\mathrm{2}\sqrt{{x}+\mathrm{4}}}={x}−\mathrm{4}\:\Rightarrow{x}=\frac{\mathrm{11}+\sqrt{\mathrm{37}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right)\:\sqrt{{x}+\mathrm{4}}={x}−\mathrm{3}+\mathrm{2}\sqrt{{x}+\mathrm{4}}\:\Rightarrow\:\sqrt{{x}+\mathrm{4}}=\mathrm{3}−{x}\:\Rightarrow \\ $$$$\:\:\:\:\:\Rightarrow\:{x}=\frac{\mathrm{7}−\sqrt{\mathrm{29}}}{\mathrm{2}}\: \\ $$$$\Rightarrow\:\mathrm{wrong} \\ $$$$\Rightarrow\:{t}=\mathrm{0}\:\Rightarrow\:{y}={x} \\ $$
Commented by jagoll last updated on 06/Mar/20
sif ^� if x =y , what the answer sir?
$$\mathrm{sif}\bar {\:}\:\mathrm{if}\:\mathrm{x}\:=\mathrm{y}\:,\:\mathrm{what}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{sir}? \\ $$
Commented by MJS last updated on 06/Mar/20
I solved it above but I don′t know what p mod q might mean if p, q ∉Z
$$\mathrm{I}\:\mathrm{solved}\:\mathrm{it}\:\mathrm{above} \\ $$$$\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{what}\:{p}\:\mathrm{mod}\:{q}\:\mathrm{might}\:\mathrm{mean} \\ $$$$\mathrm{if}\:{p},\:{q}\:\notin\mathbb{Z} \\ $$

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