Question

Answer correctly. Correct will be brainliest. Irrelative answer will be reported. Show that the root of equation $(b-c)x^{2}$$+(c-a)x^{2}$$+(a-b)=0$ are real.

Answer 1

Step-by-step explanation:

Discriminant:

${(c - a)}^{2} - 4(b - c)(a - b) \\ = {c}^{2} + {a}^{2} - 2ac - 4ab + 4 {b}^{2} + 4ac - 4bc\\ = {a}^{2} + 4 {b}^{2} + {c}^{2} - 4ab - 4bc + 2ac \\ = {(a - 2b + c)}^{2}$

Now since the square of something can never be negative, it can either br zero or positive.

So in this case discriminant is positive or zero, so, the roots are real.

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