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

$$Resoudrelesysteme$$$$\mathrm{abc}=30$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}=10$$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ac}=31$$$$$$

Answered by Frix last updated on 30/Dec/22

$$(x-a)(x-b)(x-c)=0$$$$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0$$$$x^3-10x^2+31x-30=0$$$$x_1=2$$$$x_2=3$$$$x_3=5$$$$\mathrm{with}a<b<c:a=2b=3c=5$$

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

$$thankyou$$

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

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

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