Question

If the roots of the equation (b-c)x^2+(c-a)x+(a-b)=0 are equal the a,b,c are in ?
Ap
Gp
Agp
Hp

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{(b-c)x^2+(c-a)x+(a-b)=0}$

Its roots are equal, so,

$\sf{D=0}$

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

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

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

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

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

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

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

$\sf{\implies\,a-2b+c=0}$

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

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

So, a,b,c are in AP