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\(\cfrac{1}{a+b+x}=\cfrac{1}{a}+\cfrac{1}{b}+\cfrac{1}{x} \)
বা, \(\cfrac{1}{a+b+x}-\cfrac{1}{x}=\cfrac{1}{a}+\cfrac{1}{b} \)
বা, \(\cfrac{x-(a+b+x)}{x(a+b+x)} =\cfrac{b+a}{ab} \)
বা, \(\cfrac{x-a-b-x}{x(a+b+x)} =\cfrac{a+b}{ab} \)
বা, \(\cfrac{-(a+b)}{ax+bx+x^2 }=\cfrac{a+b}{ab} \)
বা, \(\cfrac{-1}{x^2+ax+bx}=\cfrac{1}{ab} \)
বা, \(x^2+ax+bx=-ab \)
বা, \(x^2+ax+bx+ab=0\)
বা, \(x(x+a)+b(x+a)=0 \)
বা, \((x+a)(x+b)=0 \)

অর্থাৎ,হয় \((x+a)=0∴x=-a \)
নয়, \((x+b)=0∴x=-b \)
\(∴x=-a\) ও \(x=-b\) হল \(\cfrac{1}{a+b+x}=\cfrac{1}{a}+\cfrac{1}{b}+\cfrac{1}{x}\) দ্বিঘাত সমীকরনের সমাধান ।