Question

If a, b and are distinct real numbers, then show that the equation (1/x-a) + (1/x-b ) + (1/x-c) = 0 has real roots.
Where x is not equal to a, b,c​

Answer 1

Answer:

they are real roots

Step-by-step explanation:

the full explaination is in the image

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Answer 2

a, b are distinct real numbers, the equation

(1/x-a)+(1/x-b)+(1/x-c)=0 has real roots. Where x is not equal to a, b, c. The equation (1/x-a) + (1/x-b) + (1/x-c) = 0 has real roots is shown below with step-wise explanation:

- Here, (1/x-a)+(1/x-b)+(1/x-c)=0

(1/x-a)+(1/x-b)=(1/c-x)

- (x-b+x-a)/x^2+ab-x(a+b)=1/c-x

(c-x)(2x-a-b)=x^2+ab-ax-bx

2xc-ac-bc-2x^2+ax+bx=x^2+ab-ax-bx

3x^2-2ax-2bx-2cx+ab+bc+ac=0

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

- D=b^2-4ac......(1)

b=2(a+b+c)

a=3

c=ab+bc+ac

- Now put these values in (1)

D=[2(a+b+c)}^2-4(3)(ab+bc+ac)

D=4(a^2+b^2+c^2+2ab+2bc+2ac)-4(3ab+3bc+3ac)

D=4(a^2+b^2+c^2-ab-bc-ac)

- By this we will always get a positive value. As D is positive. So, it will have real roots.