Question Number 50730 by ajfour last updated on 19/Dec/18
$$\sqrt{{x}−{a}}+\sqrt{{x}−{b}}+\sqrt{{x}−{c}}+{x}\:=\:{d} \\ $$$${solve}\:{for}\:{x}. \\ $$
Answered by behi83417@gmail.com last updated on 21/Dec/18
$${after}\:{squaring}\:{and}\:{symplifing}\:{i}\:{got} \\ $$$${this}\:{equtition}: \\ $$$$\mathrm{9}\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{Ax}}^{\mathrm{3}} +\boldsymbol{\mathrm{Bx}}^{\mathrm{2}} +\boldsymbol{\mathrm{Cx}}+\boldsymbol{\mathrm{D}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{A}}=−\mathrm{24}\left({a}+{b}\right)+\mathrm{12}{d}−\mathrm{10} \\ $$$$\boldsymbol{\mathrm{B}}=\mathrm{22}\left({a}+{b}\right)^{\mathrm{2}} +\mathrm{8}\left(\mathrm{1}−\mathrm{2}{d}\right)\left({a}+{b}\right)−\mathrm{2}{d}^{\mathrm{2}} +\mathrm{10}{c}+\mathrm{1} \\ $$$$\boldsymbol{\mathrm{C}}=−\mathrm{16}\left({a}+{b}\right)^{\mathrm{3}} −\mathrm{2}\left(\mathrm{1}−\mathrm{2}{d}\right)\left({a}+{b}\right)^{\mathrm{2}} − \\ $$$$\:−\mathrm{8}\left({c}−\mathrm{4}{d}^{\mathrm{2}} \right)\left({a}+{b}\right)−\mathrm{2}{c}−\mathrm{4}{dc}−\mathrm{2}{d}^{\mathrm{2}} −\mathrm{4}{d}^{\mathrm{3}} \\ $$$$\boldsymbol{\mathrm{D}}=\left({a}+{b}\right)^{\mathrm{4}} +\mathrm{2}\left({c}−{d}^{\mathrm{2}} \right)\left({a}+{b}\right)^{\mathrm{2}} +\left({c}+{d}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$
Commented by ajfour last updated on 21/Dec/18
$${Sir},\:{please}\:{check}\:{if} \\ $$$$\:\:\:{a}\:=\:\mathrm{31},\:{b}=\:\mathrm{47},\:{c}\:=\:\mathrm{55},\:{d}=\:\mathrm{65} \\ $$$${we}\:{should}\:{get}\:\:\:{x}\:=\:\mathrm{56}\:. \\ $$