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Question Number 183835 by a.lgnaoui last updated on 30/Dec/22

Resoudre le systeme abc =30 a+b+c =10 ab+bc+ac=31

$${Resoudre}\:{le}\:{systeme} \\ $$$$\:\:\mathrm{abc}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{30} \\ $$$$\:\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:\:\:\:\:\:\:=\mathrm{10} \\ $$$$\:\:\:\mathrm{ab}+\mathrm{bc}+\mathrm{ac}=\mathrm{31} \\ $$$$ \\ $$

Answered by Frix last updated on 30/Dec/22

(x−a)(x−b)(x−c)=0 x^3 −(a+b+c)x^2 +(ab+bc+ca)x−abc=0 x^3 −10x^2 +31x−30=0 x_1 =2 x_2 =3 x_3 =5 with a<b<c: a=2 b=3 c=5

$$\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\left({a}+{b}+{c}\right){x}^{\mathrm{2}} +\left({ab}+{bc}+{ca}\right){x}−{abc}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{10}{x}^{\mathrm{2}} +\mathrm{31}{x}−\mathrm{30}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\mathrm{2} \\ $$$${x}_{\mathrm{2}} =\mathrm{3} \\ $$$${x}_{\mathrm{3}} =\mathrm{5} \\ $$$$\mathrm{with}\:{a}<{b}<{c}:\:{a}=\mathrm{2}\:{b}=\mathrm{3}\:{c}=\mathrm{5} \\ $$

Commented by a.lgnaoui last updated on 30/Dec/22

thank you

$${thank}\:{you} \\ $$

Answered by a.lgnaoui last updated on 30/Dec/22

abc=30 c=((30)/(ab)) (a+b)+((30)/(ab))=10 ab+(a+b)((30)/(ab))=31 (a+b)=10−((30)/(ab)) ab=x ⇒x+(10−((30)/x))((30)/x)=31 x+((300)/x)−((900)/x^2 )=31 x^3 −31x^2 +300x−900=0 x=10 ab=10 ⇒c=((30)/(10))=3 a+b=10−3=7 z^2 −7z+10=0 z=((7±3)/2)⇒ a=5 b=2

$${abc}=\mathrm{30}\:\:\:{c}=\frac{\mathrm{30}}{{ab}} \\ $$$$\left({a}+{b}\right)+\frac{\mathrm{30}}{{ab}}=\mathrm{10}\:\:\:\: \\ $$$${ab}+\left({a}+{b}\right)\frac{\mathrm{30}}{{ab}}=\mathrm{31} \\ $$$$\left({a}+{b}\right)=\mathrm{10}−\frac{\mathrm{30}}{{ab}} \\ $$$${ab}={x}\:\:\Rightarrow{x}+\left(\mathrm{10}−\frac{\mathrm{30}}{{x}}\right)\frac{\mathrm{30}}{{x}}=\mathrm{31} \\ $$$${x}+\frac{\mathrm{300}}{{x}}−\frac{\mathrm{900}}{{x}^{\mathrm{2}} }=\mathrm{31} \\ $$$${x}^{\mathrm{3}} −\mathrm{31}{x}^{\mathrm{2}} +\mathrm{300}{x}−\mathrm{900}=\mathrm{0} \\ $$$${x}=\mathrm{10}\:\:\:\:\:{ab}=\mathrm{10}\:\:\:\:\Rightarrow{c}=\frac{\mathrm{30}}{\mathrm{10}}=\mathrm{3} \\ $$$${a}+{b}=\mathrm{10}−\mathrm{3}=\mathrm{7}\:\:\:\: \\ $$$${z}^{\mathrm{2}} −\mathrm{7}{z}+\mathrm{10}=\mathrm{0} \\ $$$${z}=\frac{\mathrm{7}\pm\mathrm{3}}{\mathrm{2}}\Rightarrow\:\:\:\:\:{a}=\mathrm{5}\:\:\:\:\:{b}=\mathrm{2} \\ $$

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