Mathematics - Quadratic Equation Question with Solution | TestHub

Given equation is $(x-a)(x-b)+(x-b)$ $(x-c)+(x-c)(x-a)=0$ $\Rightarrow 3x^2-2(b+a+c)x+ab+bc+ca=0$ Now, here $A=3,B=-2(a+b+c)$ $C=ab+bc+ca$ $\therefore D=\sqrt{B^2-4AC}$ $=\sqrt{(-2(a+b+c){)}^2-4(3)(ab+bc+ca)}$ $=\sqrt{4(a+b+c{)}^2-12(ab+bc+ca)}$ $=2\sqrt{a^2+b^2+c^2-ab-bc-ca}$ $=2\sqrt{\frac{1}{2}\left\{(a-b{)}^2-(b-c{)}^2+(c-a{)}^2\right\}}\ge 0$