Question

For how many values of $a$, equation $(a^2-3a+2)x^2+(a^2-4)x+a^2-a-2=0$ is an identity

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$1$

For the equation $(a^2-3a+2)x^2+(a^2-4)x+a^2-a-2=0$ to be an identity in $x$, the co-efficients of $x^2$ , $x$ and the constant term must be identically zero.
$\Rightarrow a^2-3a+2=(a-1)(a-2)=0$
$\therefore a=1$ or $a=2$
$\Rightarrow a^2-4=(a-2)(a+2)=0$
$\therefore a=2$ or $a=-2$
$\Rightarrow a^2-a-2=(a+1)(a-2)=0$
$\therefore a=-1$ or $a=2$
$\therefore$ For $a=2$ all three coefficients becomes zero.
Hence, option A.