Question

(a) If the roots of the equation, $(b-c)x^2+(c-a)x+(a-b)=0$ be equal, then prove thar a,b,c are in arithmetical progression.
(b) If $a(b-c)x^2+b(c-a)x+c(a-b)=0$ has equal roots, prove that a,b,c are in harmonical progression.

Answer 1

(a) ${(c-a)}^2-4(a-b)(b-c)=0$
$c^2+a^2-2ca-4ab+4ac+4b^2-4bc=0$
or $c^2+a^2-2ca+4b^2-4b(c+a)=0$
or ${(c+a)}^2+{\left(2b\right)}^2-2.2b(c+a)=0$
or ${\left[\right(c+a)-2b]}^2=0\therefore c+a-2b=0$
or $2b=a+c$ $\therefore$ a,b,c arein A.P>
Alternate : $\sum (b-c)=0\therefore x=1$ is a root since roots are equal, therefore both the roots are 1,1.
Hence their product.
$P=1=\frac{a-b}{b-c}\therefore 2b=a+c$
$\therefore$ a,b,c are in A.P.
(b) $b^2{(c-a)}^2-4ac(b-c)(a-b)=0$
or $b^2(c^2+a^2-2ac)-4ac[ab-ac-b^2+bc]=0$
or $b^2(c^2+a^2-2ac+4ac)=4a^2{c}^2-4abc(c+a)=0$
or ${\left[b\right(c+a\left)\right]}^2+{\left(2ac\right)}^2-2.2ac.b(c+a)=0$
or ${\left[b\right(c+a)-2ac]}^2=0\therefore b(c+a)=2ac$
or $b=\frac{2ac}{a+c}$
$\therefore$ b is H.M. of a and c i.e. a,b,c are in H.P.
Alternate : Here $\sum a(b-c)=0$
$\therefore$ $x=1$ is a root and since both roots are equal, they are 1,1.
$\therefore$ $P=1=\frac{c(a-b)}{a(b-c)}$ or $b=\frac{2ac}{a+c}$
$\therefore$ a,b,c are in H.P.