How was this quadratic formed?

Question

For $[x − (a + b + c)][\frac{1}{x − a}+\frac{1}{x − b}+\frac{1}{x − c}]= 0.$

One solution is $x = a + b + c$. The notes say that the other two solutions are the roots of the quadratic equation $3x^2 − 2(a + b + c)x + (ab + bc + ac) = 0.$

I understand that Vieta's Relation has been used. But could you please elaborate how one arrives at that quadratic?

Answer 1

$$\frac1{x-a}+\frac1{x-b}+\frac1{x-c}=0$$

Make the denominator the same and multiply by the denominator: $$(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b)=0$$

$$3x^2-2(a+b+c)x+(bc+ac+ab)=0$$

Answer 2

The other solutions come from the equation : $$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0.$$

You can get rid of the denominators by multiplying both sides by $(x-a)(x-b)(x-c)$ : $$(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0.$$ Now, expand everything and you get : $$3x^2 - (b+c+a+c+a+b)x + (bc+ac+ab) = 0.$$

Hope it helps.

Answer 3

Hint:

$$\dfrac1{x-a}+\dfrac1{x-b}=-\dfrac1{x-c}$$

$$\implies(2x-a-b)(x-c)=-(x-a)(x-b)$$

$$\implies2x^2-x(a+b+2c)+ca+bc=-x^2+(a+b)x-ab$$