If a, b, c are real, then both the roots of the $$equation(x$$−$$b)(x$$−$$c)+(x$$−$$c)(x$$−$$a)+(x$$−$$a)(x$$−$$b)=0$$ $$(1)are$$ always

Question

If a, b, c are real, then both the roots of the $$equation(x$$−$$b)(x$$−$$c)+(x$$−$$c)(x$$−$$a)+(x$$−$$a)(x$$−$$b)=0$$                                  $$(1)are$$ always

Infinity Learn · Accepted Answer

We can write (1) $$as3x2$$−$$2(a+b+c)x+bc+ca+ab=0$$                                    (2)The$$ discriminant D of $$(2) is given $$byD=4(a+b+c)2$$−$$4$$×$$3(bc+ca+ab)=4a2+b2+c2+2bc+2ca+2ab$$−$$3bc$$−$$3ca$$−$$3ab=4a2+b2+c2$$−bc−ca−$$ab=2b2+c2$$−$$2bc+c2+a2$$−$$2ca+a2+b2$$−$$2ab=2(b$$−$$c)2+(c$$−$$a)2+(a$$−$$b)2As$$ a, b, c are real, we get D ≥ $$0$$. Thus, roots of (1) are real.

If $a, b, c$ are real, then both the roots of the equation
$(x - b) (x - c) + (x - c) (x - a) + (x - a) (x - b) = 0$ (1)
are always

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If a, b, c are real, then both the roots of the equation(x−b)(x−c)+(x−c)(x−a)+(x−a)(x−b)=0 (1)are always

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If $a, b, c$ are real, then both the roots of the equation
$(x - b) (x - c) + (x - c) (x - a) + (x - a) (x - b) = 0$ (1)
are always