Question

The roots of the equation $(a+c-b)x^2-2cx+(b+c-a)=0$ are

• A. $1,\frac{a+c-b}{b+c-a}$
• B. $1,\frac{b+c-a}{2c}$
• C. $1,\frac{b+c-a}{a+c-b}$
• D. $1,\frac{2c}{a+c-b}$

Answer 1

Answer: (C)

$1,\frac{b+c-a}{a+c-b}$

Given equation is $(a+c-b)x^2-2cx+(b+c-a)=0$
Now $B^2-4AC$
$=4c^2-4(c+(a-b\left)\right)(c-(a-b\left)\right)$
$=4[c^2-c^2+(a-b{)}^2]$
$=4\left[\right(a-b{)}^2]$
Hence, $x=\frac{-B\pm \sqrt{B^2-4AC}}{2A}$
$=\frac{2c\pm \sqrt{4(a-b{)}^2}}{2(a+c-b)}$
$=\frac{2c\pm 2(a-b)}{2(a+c-b)}$
$\Rightarrow x=\frac{c+a-b}{c+a-b}$ or $x=\frac{c+b-a}{c+a-b}$
Hence, $x=1$ or $x=\frac{c+b-a}{c+a-b}$