Question

If the roots of the equation $(b-c)x^2+(c-a)x+(a-b)=0$ be equal then b is equal to

• A. $\frac{a+c}{2}$
• B. $\frac{a-c}{2}$
• C. $\frac{a+c}{2ac}$
• D. $\frac{a-c}{2ac}$

Answer 1

The correct option is A $\frac{a+c}{2}$

Solution :-

Given $(b-c)x^2+(c-a)x+(a-b)=0$ have equal roots

then $D=0$

$\Rightarrow (c-a{)}^2-4(b-c\left)\right(a-b)=0$

$\Rightarrow c^2+a^2-2ac-4[ab+b^2-ac+bc]=0$

$\Rightarrow c^2+a^2-2ac-4ab-4b^2+4ac-4bc=0$

$\Rightarrow c^2+a^2+2ac-4ab-4b^2-4bc=0$

$\Rightarrow (a+b{)}^2-4b(a+b)+4b^2=0$

$\Rightarrow \left[\right(a+c)-2b{]}^2=0$

$\Rightarrow a+c=2b\Rightarrow b=\frac{a+c}{2}$