Question

please answer the question​

Answer 1

Answer:

Step-by-step explanation:

It is given that the roots of the equation

$\sf{a(b-c)\,x^2+b(c-a)\,x+c(a-b)=0}$    are equal.

So,

$\sf{b^2(c-a)^2-4\cdot\,a(b-c)\cdot\,c(a-b)=0}$

$\sf{\implies\,b^2(c^2+a^2-2ac)-4\,ac(b-c)(a-b)=0}$

$\sf{\implies\,b^2(c^2+a^2-2ac)-4\,ac(ab-ac-b^2+bc)=0}$

$\sf{\implies\,b^2c^2+a^2b^2-2ab^2c-4\,a^2bc+4c^2a^2+4ab^2c-4abc^2=0}$

$\sf{\implies\,a^2b^2+b^2c^2+4c^2a^2+2ab^2c-4abc^2-4a^2bc=0}$

$\sf{\implies\,\left(ab\right)^2+\left(bc\right)^2+\left(-2ca\right)^2+2\cdot\,ab\cdot\,bc+2\cdot\,bc\cdot\,(-2ca)+2\cdot(-2ca)\cdot\,ab=0}$

$\sf{\implies\,\left(ab+bc-2ca\right)^2=0}$

$\sf{\implies\,ab+bc-2ca=0}$

$\sf{\implies\,ab+bc=2ca}$

$\sf{\implies\,2ca=ab+bc}$

$\sf{\implies\,\dfrac{2ca}{abc}=\dfrac{ab}{abc}+\dfrac{bc}{abc}}$

$\sf{\implies\,\dfrac{2}{b}=\dfrac{1}{c}+\dfrac{1}{a}}$

$\sf{\implies\,\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}}$