If the roots of the equation $(b-c)x^2+(c-a)x+(a-b)=0$ are equal. Prove that $2b=a+c$
Given: Roots of the equation $(b-c)x^2+(c-a)x+(a-b)=0$ are equal.
To do: To prove that $2b=a+c$
Solution:
$(b-c)x^2+(c-a)x+(a-b)=0$
Comparing with quadratic equation
$Ax^2+Bx+C=0$
$A=b-c$, $B=c-a$, $C=a-b$
Discriminate when roots are equal
$\Rightarrow D=B^2-4AC=0$
$\Rightarrow D=(c-a)^2−4(b-c)(a-b)=0$
$\Rightarrow D=(c^2+a^2−2ac)-4(ba-ac-b^2+bc)=0$
$\Rightarrow D=c^2+a^2−2ac-4ab+4ac+4b^2-4bc=0$
$\Rightarrow c^2+a^2+2ac-4b(a+c)+4b^2=0$
$\Rightarrow (a+c)^2-4b(a+c)+4b^2=0$
$\Rightarrow [(a+c)-2b]^2=0$
$\Rightarrow a+c=2b$
Hence, proved.
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