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Question Number 25074 by Mr easy last updated on 03/Dec/17

Commented by moxhix last updated on 03/Dec/17

(x−a)(x−b)(x−c)(x−d)=0 x^4 −(a+b+c+d)x^3 +(ab+bc+cd+da)x^2 −(abc+bcd+cda+dab)x+abcd=0 { ((a+b+c+d=0)),((ab+bc+cd+da=0)),((abc+bcd+cda+dab=k)),((abcd=−15)) :} ((a+b+c)/d^2 )=((a+b+c+d)/d^2 )−(1/d)=−(1/d) so (x−((a+b+c)/d^2 ))(x−((b+c+d)/a^2 ))(x−((c+d+a)/b^2 ))(x−((d+a+b)/c^2 ))=0 (x+(1/d))(x+(1/a))(x+(1/b))(x+(1/c))=0 (↓×abcd≠0) (ax+1)(bx+1)(cx+1)(dx+1)=0 abcdx^4 +(abc+bcd+cda+dab)x^3 +(ab+bc+cd+da)x^2 +(a+b+c+d)x+1=0 −15x^4 +kx^3 +1=0 15x^4 −kx^3 −1=0 ∴A(15x^4 −kx^3 −1)=0 (A≠0)

$$\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)\left({x}−{d}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\left({a}+{b}+{c}+{d}\right){x}^{\mathrm{3}} +\left({ab}+{bc}+{cd}+{da}\right){x}^{\mathrm{2}} −\left({abc}+{bcd}+{cda}+{dab}\right){x}+{abcd}=\mathrm{0} \\ $$$$\begin{cases}{{a}+{b}+{c}+{d}=\mathrm{0}}\\{{ab}+{bc}+{cd}+{da}=\mathrm{0}}\\{{abc}+{bcd}+{cda}+{dab}={k}}\\{{abcd}=−\mathrm{15}}\end{cases} \\ $$$$\frac{{a}+{b}+{c}}{{d}^{\mathrm{2}} }=\frac{{a}+{b}+{c}+{d}}{{d}^{\mathrm{2}} }−\frac{\mathrm{1}}{{d}}=−\frac{\mathrm{1}}{{d}} \\ $$$${so} \\ $$$$\left({x}−\frac{{a}+{b}+{c}}{{d}^{\mathrm{2}} }\right)\left({x}−\frac{{b}+{c}+{d}}{{a}^{\mathrm{2}} }\right)\left({x}−\frac{{c}+{d}+{a}}{{b}^{\mathrm{2}} }\right)\left({x}−\frac{{d}+{a}+{b}}{{c}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$$\left({x}+\frac{\mathrm{1}}{{d}}\right)\left({x}+\frac{\mathrm{1}}{{a}}\right)\left({x}+\frac{\mathrm{1}}{{b}}\right)\left({x}+\frac{\mathrm{1}}{{c}}\right)=\mathrm{0}\:\:\:\:\:\:\left(\downarrow×{abcd}\neq\mathrm{0}\right) \\ $$$$\left({ax}+\mathrm{1}\right)\left({bx}+\mathrm{1}\right)\left({cx}+\mathrm{1}\right)\left({dx}+\mathrm{1}\right)=\mathrm{0} \\ $$$${abcdx}^{\mathrm{4}} +\left({abc}+{bcd}+{cda}+{dab}\right){x}^{\mathrm{3}} +\left({ab}+{bc}+{cd}+{da}\right){x}^{\mathrm{2}} +\left({a}+{b}+{c}+{d}\right){x}+\mathrm{1}=\mathrm{0} \\ $$$$−\mathrm{15}{x}^{\mathrm{4}} +{kx}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{15}{x}^{\mathrm{4}} −{kx}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$$\therefore{A}\left(\mathrm{15}{x}^{\mathrm{4}} −{kx}^{\mathrm{3}} −\mathrm{1}\right)=\mathrm{0}\:\:\left({A}\neq\mathrm{0}\right) \\ $$

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