2026/004) If roots of the equation $a(b−c)x^2+b(c−a)x+c(a−b)=0$ are equal, then prove that 'a', 'b', 'c' are in harmonic proportion.

 Because roots are equal so the descrminamt is zero

or $b^2(c-a)^2-4ac(b-c)(a-b) = 0$

becuase we need to show that $a,b,c$ are in HP so $\frac{1}{a} = p , \frac{1}{b} = q, \frac{1}{c} = r$ are in AP

now given relation is

$\frac{1}{q^2} (\frac{1}{r} - \frac{1}{p})^2- 4(\frac{1}{p})(\frac{1}{q})(\frac{1}{q}- \frac{1}{r}) (\frac{1}{p}- \frac{1}{q})=0 $ 

or $\frac{(p-r)^2}{p^2q^2r^2}$$ - 4(\frac{(r-q)(q-p)}{q^2r^2p^2})=0 $

or $(r-p)^2 -4 (r-q)(q-p) = 0$

Let $x= r-q$ and $y = q-p$

so we get $r - p = x + y$

or $(x+y)^2 - 4xy= 0$

or $(x-y)^2 = 0$

or $x = y$ or $q-p = r-q$ so   $\frac{1}{a} = p , \frac{1}{b} = q, \frac{1}{c} = r$ are in AP and hence $a,b,c$ are in HP