Question

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

• A. $a=2b$
• B. $a=b$
• C. $a=2c$
• D. $a=b=c$

Answer 1

Answer: (B)

$a=b$

Given the roots of the equation $(a-2b+c)x^2+(b+a-2c)x+(b+c-2a)=0$ be equal.

Then the discreminant of the equation (1) is $0$.

Then,

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

or, $(b^2+a^2+4c^2+2ab-4ac-4bc)-4(ab+ac-2a^2+bc+c^2-2ac-2b^2-2bc+4bc)=0$

or, $9a^2+9b^2-18ab=0$

or, $9(a^2+b^2-2ab)=0$

or, $(a-b{)}^2=0$

or, $a=b$.